Indices

Given,

$(125)^{\frac{1}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (125)^{\frac{1}{3}} \\[1em] \Rightarrow [(5)^3]^{\frac{1}{3}} \\[1em] \Rightarrow 5^{3 \times \frac{1}{3}} \\[1em] \Rightarrow 5.\end{aligned}$$

Hence, $(125)^{\frac{1}{3}}$ = 5.

Given,

$(8)^{\frac{2}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (8)^{\frac{2}{3}} \\[1em] \Rightarrow (2^3)^{\frac{2}{3}} \\[1em] \Rightarrow 2^{3 \times \frac{2}{3}} \\[1em] \Rightarrow 2^2 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $(8)^{\frac{2}{3}} = 4$.

Given,

$\Big(\frac{1}{5}\Big)^{-2}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{1}{5}\Big)^{-2} \\[1em] \Rightarrow 5^2 \\[1em] \Rightarrow 25.\end{aligned}$$

Hence, $\Big(\frac{1}{5}\Big)^{-2} = 25$.

Given,

$(16)^{\frac{-3}{4}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (16)^{\frac{-3}{4}} \\[1em] \Rightarrow (2^4)^{\frac{-3}{4}} \\[1em] \Rightarrow 2^{4 \times -\frac{3}{4}} \\[1em] \Rightarrow (2)^{-3} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^{3} \\[1em] \Rightarrow \frac{1}{8}.\end{aligned}$$

Hence, $(16)^{\frac{-3}{4}} = \frac{1}{8}$.

Given,

$(32)^{\frac{-4}{5}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (32)^{\frac{-4}{5}} \\[1em] \Rightarrow (2^5)^{\frac{-4}{5}} \\[1em] \Rightarrow 2^{5 \times -\frac{4}{5}} \\[1em] \Rightarrow (2)^{-4} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^{4} \\[1em] \Rightarrow \frac{1}{16}.\end{aligned}$$

Hence, $(32)^{\frac{-4}{5}} = \frac{1}{16}$.

Given,

$\Big(\frac{8}{125}\Big)^{-\frac{1}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{8}{125}\Big)^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\frac{2}{5}\Big)^3\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{3 \times -\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{-1} \\[1em] \Rightarrow \frac{5}{2}.\end{aligned}$$

Hence, $\Big(\frac{8}{125}\Big)^{-\frac{1}{3}} = \frac{5}{2}$.

Given,

$(-27)^{\frac{2}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (-27)^{\frac{2}{3}} \\[1em] \Rightarrow [(-3)^3]^{\frac{2}{3}} \\[1em] \Rightarrow (-3)^{3 \times \frac{2}{3}} \\[1em] \Rightarrow (-3)^2 \\[1em] \Rightarrow 9.\end{aligned}$$

Hence, $(-27)^{\frac{2}{3}} = 9$.

Given,

$(0.001)^{-\frac{1}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (0.001)^{-\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{1}{1000}\Big)^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\frac{1}{10}\Big)^3\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{1}{10}\Big)^{3 \times -\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{1}{10}\Big)^{-1} \\[1em] \Rightarrow 10.\end{aligned}$$

Hence, $(0.001)^{-\frac{1}{3}} = 10$.

Given,

$(0.027)^{\frac{-2}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (0.027)^{\frac{-2}{3}} \\[1em] \Rightarrow \Big(\frac{27}{1000}\Big)^{-\frac{2}{3}} \\[1em] \Rightarrow \Big[\Big(\frac{3}{10}\Big)^3\Big]^{-\frac{2}{3}} \\[1em] \Rightarrow \Big(\frac{3}{10}\Big)^{3 \times -\frac{2}{3}} \\[1em] \Rightarrow \Big(\frac{3}{10}\Big)^{-2} \\[1em] \Rightarrow \Big(\frac{10}{3}\Big)^{2} \\[1em] \Rightarrow \frac{100}{9}.\end{aligned}$$

Hence, $(0.027)^{\frac{-2}{3}} = \frac{100}{9}$.

Given,

$\Big(\frac{1}{4}\Big)^{-2} - 3 \times (8)^{\frac{2}{3}} \times 5^0 + \Big(\frac{9}{16}\Big)^{-\frac{1}{2}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{1}{4}\Big)^{-2} - 3 \times (8)^{\frac{2}{3}} \times 5^0 + \Big(\frac{9}{16}\Big)^{-\frac{1}{2}} \\[1em] \Rightarrow (4)^2 - 3 \times [(2)^3]^{\frac{2}{3}} \times 1 + \Big(\frac{16}{9}\Big)^{\frac{1}{2}} \\[1em] \Rightarrow 16 - 3 \times (2)^2 \times 1 + \Big[\Big(\frac{4}{3}\Big)^2\Big]^{\frac{1}{2}} \\[1em] \Rightarrow 16 - 12 + \frac{4}{3} \\[1em] \Rightarrow 4 + \frac{4}{3} \\[1em] \Rightarrow \frac{12 + 4}{3} \\[1em] \Rightarrow \frac{16}{3} = 5\frac{1}{3}.\end{aligned}$$

Hence, $\Big(\frac{1}{4}\Big)^{-2} - 3 \times (8)^{\frac{2}{3}} \times 5^0 + \Big(\frac{9}{16}\Big)^{-\frac{1}{2}} = 5\frac{1}{3}$.

Given,

$\sqrt{\frac{1}{4}} + (0.01)^{\frac{-1}{2}} - (27)^{\frac{2}{3}} \times 3^0$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \sqrt{\frac{1}{4}} + (0.01)^{\frac{-1}{2}} - (27)^{\frac{2}{3}} \times 3^0 \\[1em] \Rightarrow \frac{1}{2} + \Big(\frac{1}{100}\Big)^{\frac{-1}{2}} - [(3)^3]^{\frac{2}{3}} \times 1 \\[1em] \Rightarrow \frac{1}{2} + (100)^{\frac{1}{2}} - (3)^2 \\[1em] \Rightarrow \frac{1}{2} + [(10)^2]^{\frac{1}{2}} - 9 \\[1em] \Rightarrow \frac{1}{2} + 10 - 9 \\[1em] \Rightarrow \frac{1}{2} + 1 \\[1em] \Rightarrow \frac{1 + 2}{2} \\[1em] \Rightarrow \frac{3}{2} \\[1em] \Rightarrow 1\frac{1}{2}.\end{aligned}$$

Hence, $\sqrt{\frac{1}{4}} + (0.01)^{\frac{-1}{2}} - (27)^{\frac{2}{3}} \times 3^0 = 1\frac{1}{2}$.

Given,

$\Big(\frac{81}{16}\Big)^{-\frac{3}{4}} \times \Big[\Big(\frac{25}{9}\Big)^{-\frac{3}{2}} ÷ {\Big(\frac{5}{2}\Big)^{-3}}\Big]$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{81}{16}\Big)^{-\frac{3}{4}} \times \Big[\Big(\frac{25}{9}\Big)^{-\frac{3}{2}} ÷ {\Big(\frac{5}{2}\Big)^{-3}}\Big] \\[1em] \Rightarrow \Big(\frac{16}{81}\Big)^{\frac{3}{4}} \times \Big[\Big(\frac{9}{25}\Big)^{\frac{3}{2}} ÷ {\Big(\frac{2}{5}\Big)^{3}}\Big] \\[1em] \Rightarrow \Big[\Big(\frac{2}{3}\Big)^4\Big]^{\frac{3}{4}} \times \Big[\Big[\Big(\frac{3}{5}\Big)^2\Big]^{\frac{3}{2}} ÷ {\Big(\frac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\frac{2}{3}\Big)^3 \times \Big[\Big(\frac{3}{5}\Big)^3 ÷ {\Big(\frac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\frac{8}{27}\Big) \times \Big[\Big(\frac{27}{125}\Big) ÷ {\Big(\frac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\frac{8}{27}\Big) \times \Big[\Big(\frac{27}{125}\Big) \times {\Big(\frac{125}{8}\Big)}\Big] \\[1em] \Rightarrow \Big(\frac{8}{27}\Big) \times \Big(\frac{27}{8}\Big) \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\Big(\frac{81}{16}\Big)^{-\frac{3}{4}} \times \Big[\Big(\frac{25}{9}\Big)^{-\frac{3}{2}} ÷ {\Big(\frac{5}{2}\Big)^{-3}}\Big] = 1$.

Given,

$\Big[(64)^{\frac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\frac{1}{2}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big[[(4)^3]^{\frac{2}{3}} \times \Big(\frac{1}{2}\Big)^{2} ÷ 1\Big]^{-\frac{1}{2}} \\[1em] \Rightarrow \Big[(4)^{3 \times \frac{2}{3}} \times \Big(\frac{1}{2}\Big)^{2} ÷ 1\Big]^{-\frac{1}{2}} \\[1em] \Rightarrow \Big[(4)^2 \times \Big(\frac{1}{4}\Big) ÷ 1\Big]^{-\frac{1}{2}} \\[1em] \Rightarrow \Big[16 \times \Big(\frac{1}{4}\Big)\Big]^{-\frac{1}{2}} \\[1em] \Rightarrow (4)^{-\frac{1}{2}} \\[1em] \Rightarrow \Big(\frac{1}{4}\Big)^{\frac{1}{2}} \\[1em] \Rightarrow \Big[\Big(\frac{1}{2}\Big)^2\Big]^{\frac{1}{2}} \\[1em] \Rightarrow \frac{1}{2}.\end{aligned}$$

Hence, $\Big[(64)^{\frac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\frac{1}{2}} = \frac{1}{2}$.

Given,

$(81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + (8)^{\frac{1}{3}} . \Big(\frac{1}{2}\Big)^{-1} . (2)^0$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + (8)^{\frac{1}{3}} \times \Big(\frac{1}{2}\Big)^{-1} \times (2)^0 \\[1em] \Rightarrow [(3)^4]^{\frac{3}{4}} - (32)^{\frac{2}{5}} + [(2)^3]^{\frac{1}{3}} \times (2) \times 1 \\[1em] \Rightarrow (3)^3 - [(2)^5]^{\frac{2}{5}} + 2^1 \times 2 \\[1em] \Rightarrow 27 - (2)^2 + 4 \\[1em] \Rightarrow 31 - 4 \\[1em] \Rightarrow 27.\end{aligned}$$

Hence, $(81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + (8)^{\frac{1}{3}} \times \Big(\frac{1}{2}\Big)^{-1} \times (2)^0 = 27$.

Given,

$\Big(\frac{16}{81}\Big)^{-\frac{3}{4}} \times \Big(\frac{49}{9}\Big)^{\frac{3}{2}} ÷ \Big(\frac{343}{216}\Big)^{\frac{2}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{16}{81}\Big)^{-\frac{3}{4}} \times \Big(\frac{49}{9}\Big)^{\frac{3}{2}} ÷ \Big(\frac{343}{216}\Big)^{\frac{2}{3}} \\[1em] \Rightarrow \Big(\frac{81}{16}\Big)^{\frac{3}{4}} \times \Big[\Big(\frac{7}{3}\Big)^2\Big]^{\frac{3}{2}} ÷ \Big[\Big(\frac{7}{6}\Big)^3\Big]^{\frac{2}{3}} \\[1em] \Rightarrow \Big[\Big(\frac{3}{2}\Big)^4\Big]^{\frac{3}{4}} \times \Big[\Big(\frac{7}{3}\Big)^2\Big]^{\frac{3}{2}} ÷ \Big[\Big(\frac{7}{6}\Big)^3\Big]^{\frac{2}{3}} \\[1em] \Rightarrow \Big(\frac{3}{2}\Big)^3 \times \Big(\frac{7}{3}\Big)^3 ÷ \Big(\frac{7}{6}\Big)^2 \\[1em] \Rightarrow \Big(\frac{27}{8}\Big) \times \Big[\Big(\frac{343}{27}\Big) ÷ \Big(\frac{49}{36}\Big)\Big] \\[1em] \Rightarrow \Big(\frac{27}{8}\Big) \times \Big[\Big(\frac{343}{27}\Big) \times \Big(\frac{36}{49}\Big)\Big] \\[1em] \Rightarrow \Big(\frac{27}{8}\Big) \times \Big(\frac{7}{3}\Big) \times 4 \\[1em] \Rightarrow \Big(\frac{9}{2}\Big) \times 7 \\[1em] \Rightarrow \frac{63}{2} = 31\frac{1}{2}.\end{aligned}$$

Hence, $\Big(\frac{16}{81}\Big)^{-\frac{3}{4}} \times \Big(\frac{49}{9}\Big)^{\frac{3}{2}} ÷ \Big(\frac{343}{216}\Big)^{\frac{2}{3}} = 31\frac{1}{2}$.

Given,

$\Big(\frac{64}{125}\Big)^{-\frac{2}{3}} ÷ \frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}} + \Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{64}{125}\Big)^{-\frac{2}{3}} ÷ \frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}} + \Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 \\[1em] \Rightarrow \Big(\frac{125}{64}\Big)^{\frac{2}{3}} ÷ \frac{1}{\Big(\frac{4^4}{5^4}\Big)^{\frac{1}{4}}} + 1 \\[1em] \Rightarrow \Big(\frac{5^3}{4^3}\Big)^{\frac{2}{3}} ÷ \frac{1}{\Big(\frac{4}{5}\Big)^{4 \times \frac{1}{4}}} + 1 \\[1em] \Rightarrow \Big(\frac{5}{4}\Big)^{3 \times \frac{2}{3}} ÷ \frac{1}{\Big(\frac{4}{5}\Big)} + 1 \\[1em] \Rightarrow \Big(\frac{5}{4}\Big)^2 ÷ \frac{5}{4} + 1 \\[1em] \Rightarrow \frac{25}{16} ÷ \frac{5}{4} + 1 \\[1em] \Rightarrow \frac{25}{16} \times \frac{4}{5} + 1 \\[1em] \Rightarrow \frac{5}{4} + 1 \\[1em] \Rightarrow \frac{5 + 4}{4} \\[1em] \Rightarrow \frac{9}{4} \\[1em] \Rightarrow 2\frac{1}{4}.\end{aligned}$$

Hence, $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}} ÷ \frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}} + \Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 = 2\frac{1}{4}$.

Given,

$\frac{(32)^{\frac{2}{5}} \times (4)^{-\frac{1}{2}} \times (8)^{\frac{1}{3}}}{2^{-2} ÷ (64)^{\frac{-1}{3}}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{[(2)^5]^{\frac{2}{5}} \times [(2)^2]^{-\frac{1}{2}} \times {[(2)^3]^{\frac{1}{3}}}}{\Big(\frac{1}{2}\Big)^{2} ÷ (4^3)^{-\frac{1}{3}}} \\[1em] \Rightarrow \frac{(2)^2 \times 2^{-1} \times 2^1}{\Big(\frac{1}{2}\Big)^{2} ÷ 4^{-1}} \\[1em] \Rightarrow \frac{(2)^2 \times \frac{1}{2} \times 2}{\Big(\frac{1}{4}\Big) ÷ \Big(\frac{1}{4}\Big)} \\[1em] \Rightarrow \frac{4 \times \frac{1}{2} \times 2}{\Big(\frac{1}{4}\Big) \times 4} \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\frac{(32)^{\frac{2}{5}} \times (4)^{-\frac{1}{2}} \times (8)^{\frac{1}{3}}}{2^{-2} ÷ (64)^{\frac{-1}{3}}} = 4$.

Given,

$(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0 \\[1em] \Rightarrow [(3)^3]^{\frac{4}{3}} + [(2)^5]^{0.8} + \Big(\frac{8}{10}\Big)^{-1} + 1 \\[1em] \Rightarrow (3)^{3 \times \frac{4}{3}} + [(2)^5]^{0.8} + \Big(\frac{8}{10}\Big)^{-1} + 1 \\[1em] \Rightarrow (3)^4 + (2)^4 + \Big(\frac{10}{8}\Big)^{1} + 1 \\[1em] \Rightarrow 81 + 16 + \frac{10}{8} + 1 \\[1em] \Rightarrow 97 + 1.25 + 1 \\[1em] \Rightarrow 99.25\end{aligned}$$

Hence, $(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0 = 99.25$.

Given,

$\Big[\frac{(27)^{-3}}{9^{-3}}\Big]^{\frac{1}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big[\frac{(27)^{-3}}{9^{-3}}\Big]^{\frac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\frac{27}{9}\Big)^{-3}\Big]^{\frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{27}{9}\Big)^{-3 \times \frac{1}{3}} \\[1em] \Rightarrow \Big(\frac{27}{9}\Big)^{-1} \\[1em] \Rightarrow \frac{9}{27} \\[1em] \Rightarrow \frac {1}{3}.\end{aligned}$$

Hence, $\Big[\frac{(27)^{-3}}{9^{-3}}\Big]^{\frac{1}{3}} = \frac{1}{3}$.

Given,

$(\sqrt{32}-\sqrt{5})^{\frac{1}{3}} (\sqrt{32}+\sqrt{5})^{\frac{1}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow [(\sqrt{32}-\sqrt{5})(\sqrt{32}+\sqrt{5})]^{\frac{1}{3}} \\[1em] \Rightarrow [(\sqrt{32})^2-(\sqrt{5})^2]^{\frac{1}{3}} \\[1em] \Rightarrow [32 - 5]^{\frac{1}{3}} \\[1em] \Rightarrow [27]^{\frac{1}{3}} \\[1em] \Rightarrow (3^3)^{\frac{1}{3}} \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $$\begin{aligned}(\sqrt{32}-\sqrt{5})^{\frac{1}{3}} \\ (\sqrt{32}+\sqrt{5})^{\frac{1}{3}} = 3\end{aligned}$$.

Given,

$(9)^{\frac{5}{2}} - 3 \times (4)^0 - \Big(\frac{1}{81}\Big)^{-\frac{1}{2}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (9)^{\frac{5}{2}} - 3(4)^0 - \Big(\frac{1}{81}\Big)^{-\frac{1}{2}} \\[1em] \Rightarrow [(3)^2]^{\frac{5}{2}} - 3(1) - (81)^\frac{1}{2} \\[1em] \Rightarrow 3^5 - 3 - (9^2)^{\frac{1}{2}} \\[1em] \Rightarrow (3)^5 - 3 - 9 \\[1em] \Rightarrow 243 - 3 - 9 \\[1em] \Rightarrow 231.\end{aligned}$$

Hence, $(9)^{\frac{5}{2}} - 3.(4)^0 - \Big(\frac{1}{81}\Big)^{-\frac{1}{2}} = 231$.

Given,

$\frac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} \\[1em] \Rightarrow \frac{3^n \times (3^2)^{n + 1}}{3^{n - 1} \times (3^2)^{n - 1}} \\[1em] \Rightarrow \frac{3^n \times (3)^{2n + 2}}{3^{n - 1} \times (3)^{2n - 2}} \\[1em] \Rightarrow \frac{3^{n + 2n + 2}}{3^{n - 1 + 2n - 2}} \\[1em] \Rightarrow \frac{3^{3n + 2}}{3^{3n - 3}} \\[1em] \Rightarrow 3^{(3n + 2) - (3n - 3)} \\[1em] \Rightarrow 3^{(3n + 2 - 3n + 3)} \\[1em] \Rightarrow 3^5.\end{aligned}$$

Hence, $\frac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} = 3^5$.

Given,

$\frac{(27)^{\frac{2n}{3}} \times (8)^{-\frac{n}{6}}}{(18)^{-\frac{n}{2}}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{[(3)^3]^{\frac{2n}{3}} \times [(2)^3]^{-\frac{n}{6}}}{[(3)^2 \times 2]^{-\frac{n}{2}}} \\[1em] \Rightarrow \frac{[(3)^{2n}] \times (2)^{-\frac{n}{2}}}{[(3)^2]^{-\frac{n}{2}} \times (2)^{-\frac{n}{2}}} \\[1em] \Rightarrow \frac{(3)^{2n}}{3^{-n}} \\[1em] \Rightarrow (3)^{2n - (-n)} \\[1em] \Rightarrow (3)^{2n + n} \\[1em] \Rightarrow 3^{3n}.\end{aligned}$$

Hence, $\frac{(27)^{\frac{2n}{3}} \times (8)^{-\frac{n}{6}}}{(18)^{-\frac{n}{2}}} = 3^{3n}$.

Given,

$\frac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}} \\[1em] \Rightarrow \frac{5^{2n + 12} \times [(5)^2]^{-7 + 2n}}{[(5)^3]^{2n}} \\[1em] \Rightarrow \frac{5^{2n + 12} \times 5^{2(-7 + 2n)}}{(5)^{6n}} \\[1em] \Rightarrow \frac{5^{2n + 12} \times 5^{-14 + 4n}}{(5)^{6n}} \\[1em] \Rightarrow \frac{5^{2n + 12 + (-14 + 4n)}}{(5)^{6n}} \\[1em] \Rightarrow \frac{5^{6n - 2}}{(5)^{6n}} \\[1em] \Rightarrow 5^{6n - 2 - 6n} \\[1em] \Rightarrow 5^{-2} \\[1em] \Rightarrow \Big(\frac{1}{5}\Big)^{2} \\[1em] \Rightarrow \frac{1}{25}.\end{aligned}$$

Hence, $\frac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}} = \frac{1}{25}$.

Given,

$\frac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} \\[1em] \Rightarrow \frac{5^{n + 2 + 1} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} \\[1em] \Rightarrow \frac{5^{n + 1} \times 5^{2} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n} \times 5 ^{1}} \\[1em] \Rightarrow \frac{5^{n + 1}(5^2 - 16)}{5^n (12 - 2 \times 5)} \\[1em] \Rightarrow \frac{5^{n + 1 - n}(25 - 16)}{(12 - 10)} \\[1em] \Rightarrow \frac{5 \times 9}{(2)} \\[1em] \Rightarrow \frac{45}{2} = 22.5\end{aligned}$$

Hence, $\frac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} = 22.5$.

Given,

$\frac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n} \\[1em] \Rightarrow \frac{3 \times [(3)^3]^{n + 1} + 3^2 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times [(3)^3]^n} \\[1em] \Rightarrow \frac{3 \times 3^{3n + 3} + 3^2 \times 3^{3n - 1}}{8 \times 3^n - 5 \times 3^n} \\[1em] \Rightarrow \frac{(3)^{3n + 3 + 1} + 3^{(3n - 1 + 2)}}{8 \times 3^{3n} - 5 \times (3)^{3n}} \\[1em] \Rightarrow \frac{(3)^{3n + 1 + 3} + 3^{(3n + 1)}}{3^{3n}(8- 5)} \\[1em] \Rightarrow \frac{(3)^{3n + 1} \times 3^3 + 3^{(3n + 1)}}{3^{3n}.3^1} \\[1em] \Rightarrow \frac{3^{3n + 1}.(3^3 + 1)}{3^{3n + 1}} \\[1em] \Rightarrow \frac{(3)^{3n + 1}.(27 + 1)}{3^{3n + 1}} \\[1em] \Rightarrow 28.\end{aligned}$$

Hence, $\frac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n} = 28$.

Given,

$\frac{5 \times (25)^{n + 1} - 25 \times 5^{2n}}{5 \times 5^{(2n + 3)} - (25)^{n + 1}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{5 \times (25)^{n + 1} - 5^2 \times 5^{2n}}{5^1 \times 5^{2n + 3} - (25)^{n + 1}} \\[1em] \Rightarrow \frac{5 \times [(5)^2]^{n + 1} - 5^{2n + 2}}{5^{2n + 3 + 1} - [(5)^2]^{n + 1}} \\[1em] \Rightarrow \frac{5^1 \times (5)^{2n + 2} - 5^{2n + 2}}{5^{2n + 4} - (5)^{2n + 2}} \\[1em] \Rightarrow \frac{5^{2n + 2 + 1} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] \Rightarrow \frac{5^{2n + 3} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] \Rightarrow \frac{5^{2n + 2}(5 - 1)}{5^{2n + 2}(5^2 - 1)} \\[1em] \Rightarrow \frac{4}{(25 - 1)} \\[1em] \Rightarrow \frac{4}{24} \\[1em] \Rightarrow \frac{1}{6}.\end{aligned}$$

Hence, $\frac{5 \times (25)^{n + 1} - 25 \times 5^{2n}}{5 \times 5^{2n + 3} - (25)^{n + 1}} = \frac{1}{6}$.

Given,

$\frac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\frac{2}{3}}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{(2n + 3)} - [(7)^2]^{n + 2}}{\Big[[(7)^3]^{n + 1}\Big]^{\frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{(2n + 3)} - (7)^{2n + 4}}{7^{3 \times (n + 1) \times \frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{(2n + 3)} - 7^{2n + 3 + 1}}{7^{{(n + 1)} \times 2}} \\[1em] \Rightarrow \frac{7^{(2n + 3)}(1 - 7)}{(7)^{2n + 2}} \\[1em] \Rightarrow 7^{(2n + 3) - (2n + 2)} \times -6 \\[1em] \Rightarrow 7^{2n + 3 - 2n - 2} \times -6 \\[1em] \Rightarrow 7^1 \times -6 \\[1em] \Rightarrow -42.\end{aligned}$$

Hence, $\frac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\frac{2}{3}}} = - 42$.

Given,

$\Big(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\Big)(x^{\frac{2}{3}} + 1 + x^{-\frac{2}{3}}\Big)$

Simplifying the expression:

$$\begin{aligned}\Rightarrow \Big(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\Big)(x^{\frac{2}{3}} + 1 + x^{-\frac{2}{3}}\Big) \\[1em] \Rightarrow \Big(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\Big) \Big[(x^{\frac{1}{3}})^2 + x^{\frac{1}{3}} ⋅ x^{-\frac{1}{3}} + (x^{-\frac{1}{3}})^2\Big]\end{aligned}$$

Simplifying the expression using the identity,

(a − b)(a² + ab + b²) = a³ − b³

$$\begin{aligned}\Rightarrow \Big([x^{\frac{1}{3}}]^3 - [x^{-\frac{1}{3}}]^3\Big) \\[1em] \Rightarrow \Big(x^{\frac{1}{3} \times 3} - x^{-\frac{1}{3} \times 3}\Big) \\[1em] \Rightarrow (x - x^{-1}) \\[1em] \Rightarrow x - \frac{1}{x}\end{aligned}$$

Hence, $\Big(x^{1/3} - x^{-1/3}\Big)(x^{2/3} + 1 + x^{-2/3}\Big) = x - \frac{1}{x}$.

Given,

a⁷ × a⁴ × a^-6 × a⁰

Simplifying the expression :

⇒ a^{7 + 4 + (-6)} × 1

⇒ a⁵ × 1

⇒ a⁵.

Hence, a⁷ × a⁴ × a^-6 × a⁰ = a⁵.

Given,

$a^{\frac{4}{3}} ÷ a^{\frac{-2}{3}}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow a^{\frac{4}{3}} ÷ a^{\frac{-2}{3}} \\[1em] \Rightarrow \frac{a^{\frac{4}{3}}}{a^{\frac{-2}{3}}} \\[1em] \Rightarrow a^{\frac{4}{3} - \Big(-\frac{2}{3}\Big)} \\[1em] \Rightarrow a^{\frac{4 - (-2)}{3}} \\[1em] \Rightarrow a^{\frac{4 + 2}{3}} \\[1em] \Rightarrow a^{\frac{6}{3}} \\[1em] \Rightarrow a^2.\end{aligned}$$

Hence, $a^{\frac{4}{3}} ÷ a^{\frac{-2}{3}}= a^2$.

Given,

(a^-1 + b^-1) ÷ (a^-2 - b^-2)

Simplifying the expression :

$$\begin{aligned}\Rightarrow (a^{-1} + b^{-1}) ÷ (a^{-2} - b^{-2}) \\[1em] \Rightarrow \Big(\frac{1}{a} + \frac{1}{b}\Big) ÷ \Big(\frac{1}{a^2} - \frac{1}{b^2}\Big) \\[1em] \Rightarrow \Big(\frac{b + a}{ab}\Big) ÷ \Big(\frac{b^2 - a^2}{a^2b^2}\Big) \\[1em] \Rightarrow \Big(\frac{b + a}{ab}\Big) \times \frac{a^2b^2}{b^2 - a^2}\\[1em] \Rightarrow \frac{ab(b + a)}{(b + a)(b - a)} \\[1em] \Rightarrow \frac{ab}{(b - a)}.\end{aligned}$$

Hence, $(a^{-1} + b^{-1}) ÷ (a^{-2} - b^{-2}) = \frac{ab}{(b - a)}$.

Given,

(a^-1 + b^-1) ÷ (ab)^-1

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{1}{a} + \frac{1}{b}\Big) ÷ \Big(\frac{1}{ab}\Big) \\[1em] \Rightarrow \Big(\frac{b + a}{ab}\Big) ÷ \Big(\frac{1}{ab}\Big) \\[1em] \Rightarrow \Big(\frac{a + b}{ab}\Big) \times ab \\[1em] \Rightarrow a + b.\end{aligned}$$

Hence, $(a^{-1} + b^{-1}) ÷ (ab)^{-1} = (a + b)$.

Given,

(a^-1 × b^-1) ÷ (a^-1 + b^-1)

Simplifying the expression :

$$\begin{aligned}\Rightarrow (a^{-1} \times b^{-1}) ÷ (a^{-1} + b^{-1}) \\[1em] \Rightarrow \Big(\frac{1}{a} \times \frac{1}{b}\Big) ÷ \Big(\frac{1}{a} + \frac{1}{b}\Big) \\[1em] \Rightarrow \frac{1}{ab} ÷ \frac{b + a}{ab} \\[1em] \Rightarrow \frac{1}{ab} \times \frac{ab}{a + b} \\[1em] \Rightarrow \frac{1}{a + b}.\end{aligned}$$

Hence, $a^{-1} \times b^{-1} ÷ a^{-1} + b^{-1} = \frac{1}{a + b}$.

Given,

(a + b)^-1 × (a^-1 + b^-1)

Simplifying the expression :

$$\begin{aligned}\Rightarrow (a + b)^{-1} \times (a^{-1} + b^{-1}) \\[1em] \Rightarrow \Big(\frac{1}{a + b}\Big) \times \Big(\frac{1}{a} + \frac{1}{b}\Big) \\[1em] \Rightarrow \Big(\frac{1}{a + b}\Big) \times \Big(\frac{b + a}{ab}\Big) \\[1em] \Rightarrow \frac{1}{ab}.\end{aligned}$$

Hence, $(a + b)^{-1} \times (a{-1} + b{-1}) = \frac{1}{ab}$.

Given,

$\frac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})}$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} \\[1em] \Rightarrow \frac{(a + b +c)}{\Big(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\Big)} \\[1em] \Rightarrow \frac{(a + b + c)}{\Big(\frac{c + a + b}{abc}\Big)} \\[1em] \Rightarrow {\Big(\frac{(a + b + c) \times abc}{a + b + c}\Big)} \\[1em] \Rightarrow abc.\end{aligned}$$

Hence, $\frac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} = abc$.

Given,

$\Big(\frac{x^a}{x^b}\Big)^{a + b} \times \Big(\frac{x^b}{x^c}\Big)^{b + c} \times \Big(\frac{x^c}{x^a}\Big)^{c + a} = 1$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^b}\Big)^{a + b} \times \Big(\frac{x^b}{x^c}\Big)^{b + c} \times \Big(\frac{x^c}{x^a}\Big)^{c + a} \\[1em] \Rightarrow (x^{a - b})^{a + b} \times (x^{b - c})^{b + c} \times (x^{c - a})^{c + a} \\[1em] \Rightarrow (x^{a^2 - b^2}) \times (x^{b^2 - c^2}) \times (x^{c^2 - a^2}) \\[1em] \Rightarrow (x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}) \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\Big(\frac{x^a}{x^b}\Big)^{a + b} \times \Big(\frac{x^b}{x^c}\Big)^{b + c} \times \Big(\frac{x^c}{x^a}\Big)^{c + a} = 1$.

Given,

$\Big(\frac{x^a}{x^b}\Big)^{\frac{1}{ab}} \times \Big(\frac{x^b}{x^c}\Big)^{\frac{1}{bc}} \times \Big(\frac{x^c}{x^a}\Big)^{\frac{1}{ac}} = 1$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^b}\Big)^{\frac{1}{ab}} \times \Big(\frac{x^b}{x^c}\Big)^{\frac{1}{bc}} \times \Big(\frac{x^c}{x^a}\Big)^{\frac{1}{ac}} \\[1em] \Rightarrow (x^{a - b})^{\frac{1}{ab}} \times (x^{b - c})^{\frac{1}{bc}} \times (x^{c - a})^{\frac{1}{ca}} \\[1em] \Rightarrow (x)^{\frac{a - b}{ab}} \times (x)^{\frac{b - c}{bc}} \times (x)^{\frac{c - a}{ac}} \\[1em] \Rightarrow (x)^{\frac{a}{ab}- \frac{b}{ab}} \times (x)^{\frac{b}{bc} -\frac{c}{bc}} \times (x)^{\frac{c}{ac}- \frac{a}{ac}} \\[1em] \Rightarrow \Big[(x)^{\frac{1}{b} - \frac{1}{a}}\Big] \times \Big[(x)^{\frac{1}{c} - \frac{1}{b}}\Big] \times \Big[(x)^{\frac{1}{a} - \frac{1}{c}}\Big] \\[1em] \Rightarrow \Big[(x)^{\frac{1}{b} - \frac{1}{a} + \frac{1}{c} - \frac{1}{b} + \frac{1}{a} - \frac{1}{c}}\Big] \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\Big(\frac{x^a}{x^b}\Big)^{ab} \times \Big(\frac{x^b}{x^c}\Big)^{bc} \times \Big(\frac{x^c}{x^a}\Big)^{ca} = 1$.

Given,

$\Big(\frac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\frac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\frac{x^c}{x^a}\Big)^{c + a - b} = 1$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\frac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\frac{x^c}{x^a}\Big)^{c + a - b} \\[1em] \Rightarrow x^{(a - b)(a + b - c)} \times x^{(b - c)(b + c - a)} \times x^{(c - a)(c + a - b)}\\[1em] \Rightarrow x^{(a^2 + ab − ac − ab − b^2 + bc)} \times x^{(b^2 + bc − ab − bc − c^2 + ac)} \times x^{(c^2 + ac − bc − ac − a^2 + ab)} \\[1em] \Rightarrow x^{(a^2 − b^2 + bc − ac)} \times x^{(b^2 − c^2 + ac − ab)} \times x^{(c^2 − a^2 + ab − bc)} \\[1em] \Rightarrow x^{(a^2 − b^2 + bc − ac) + (b^2 − c^2 + ac − ab) + (c^2 − a^2 + ab − bc)} \\[1em] \Rightarrow x^{(a^2 − a^2 − b^2 + b^2 − c^2 + c^2 + bc − bc − ac + ac − ab + ab )} \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\Big(\frac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\frac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\frac{x^c}{x^a}\Big)^{c + a - b} = 1$.

Given,

$\frac{a^{-1}}{(a^{-1} + b^{-1})} + \frac{a^{-1}}{(a^{-1} - b^{-1})} = \frac{2b^2}{(b^2 - a^2)}$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \frac{a^{-1}}{(a^{-1} + b^{-1})} + \frac{a^{-1}}{(a^{-1} - b^{-1})} \\[1em] \Rightarrow \frac{\frac{1}{a}}{\Big(\frac{1}{a} + \frac{1}{b}\Big)} + \frac{\frac{1}{a}}{\Big(\frac{1}{a} - \frac{1}{b}\Big)} \\[1em] \Rightarrow \frac{1}{a} \times \frac{1}{\Big(\frac{b + a}{ab}\Big)} + \frac{1}{a} \times \frac{1}{\Big(\frac{b - a}{ab}\Big)} \\[1em] \Rightarrow \frac{1}{a} \times \frac{ab}{b + a} + \frac{1}{a} \times \frac{ab}{b - a} \\[1em] \Rightarrow \frac{b}{b + a} + \frac{b}{b - a} \\[1em] \Rightarrow \frac{b(b - a) + b(b + a)}{(b)^2 - (a)^2} \\[1em] \Rightarrow \frac{b^2 - ba + b^2 + ba}{b^2 - a^2} \\[1em] \Rightarrow \frac{2b^2}{b^2 - a^2}.\end{aligned}$$

Hence proved, $\frac{a^{-1}}{(a^{-1} + b^{-1})} + \frac{a^{-1}}{(a^{-1} - b^{-1})} = \frac{2b^2}{(b^2 - a^2)}$.

Given,

$\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$

Solving L.H.S :

Multiplying numerator and denominator of first term by x^a, second term by x^b, third term by x^c we get,

$$\begin{aligned}\Rightarrow \frac{1 \times x^a}{(1 + x^{b - a} + x^{c - a})\times x^a} + \frac{1 \times x^b}{(1 + x^{a - b} + x^{c - b})\times x^b} + \frac{1 \times x^c}{(1 + x^{b - c} + x^{a - c})\times x^c} \\[1em] \Rightarrow \frac{1 \times x^a}{(1 \times x^a + x^{b - a} \times x^a + x^{c - a} \times x^a)} + \frac{1 \times x^b}{(1 \times x^b + x^{a - b} \times x^b + x^{c - b} \times x^b)} + \frac{1 \times x^c}{(1 \times x^c + x^{b - c} \times x^c + x^{a - c} \times x^c)} \\[1em] \Rightarrow \frac{x^a}{(x^a + x^{b - a + a} + x^{c - a + a})} + \frac{x^b}{(x^b + x^{a - b + b} + x^{c - b + b})} + \frac{x^c}{(x^c + x^{b - c + c} + x^{a - c + c})} \\[1em] \Rightarrow \frac{x^a}{(x^a + x^b + x^c)} + \frac{x^b}{(x^b + x^a + x^c)} + \frac{x^c}{(x^c + x^b + x^a)} \\[1em] \Rightarrow \frac{x^a + x^b + x^c}{(x^a + x^b + x^c)} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$.

Given,

$\frac{1}{1 + a + b^{-1}} + \frac{1}{1 + b + c^{-1}} + \frac{1}{1 + c + a^{-1}} = 1$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \frac{1}{1 + a + b^{-1}} + \frac{1}{1 + b + c^{-1}} + \frac{1}{1 + c + a^{-1}} \\[1em] \Rightarrow \frac{1}{1 + a + \frac{1}{b}} + \frac{1}{1 + b + \frac{1}{c}} + \frac{1}{1 + c + \frac{1}{a}}\end{aligned}$$

Multiplying numerator and denominator of first term by b, second term by c, and third term by a we get,

$$\begin{aligned}\Rightarrow \frac{1 \times b}{\Big(1 + a + \frac{1}{b}\Big) \times b} + \frac{1 \times c}{\Big(1 + b + \frac{1}{c}\Big) \times c} + \frac{1 \times a}{\Big(1 + c + \frac{1}{a}\Big) \times a} \\[1em] \Rightarrow \frac{b}{(b + ab + 1)} + \frac{c}{(c + bc + 1)} + \frac{a}{(a + ac + 1)}\end{aligned}$$

Since abc = 1, c = $\frac{1}{ab}$.

Substituting, c = $\frac{1}{ab}$, we get :

$$\begin{aligned}\Rightarrow \frac{b}{(b + ab + 1)} + \frac{\frac{1}{ab}}{\Big(\frac{1}{ab} + b \Big(\frac{1}{ab}\Big) + 1\Big)} + \frac{a}{(a + a \Big(\frac{1}{ab}\Big) + 1)} \\[1em] \Rightarrow \frac{b}{(b + ab + 1)} + \frac{\frac{1}{ab}}{\Big(\frac{1}{ab} + \frac{1}{a} + 1\Big)} + \frac{a}{\Big(a + \frac{1}{b} + 1\Big)} \\[1em] \Rightarrow \frac{b}{(b + ab + 1)} + \frac{1}{ab \times \Big(\frac{1 + b + ab}{ab}\Big)} + \frac{a}{\Big(\frac{ab + 1 + b}{b}\Big)} \\[1em] \Rightarrow \frac{b}{(b + ab + 1)} + \frac{1}{\Big({1 + b + ab}\Big)} + \frac{ab}{\Big(ab + 1 + b\Big)} \\[1em] \Rightarrow \frac{b + 1 + ab}{(b + ab + 1)} \\[1em] \Rightarrow \frac{1 + b + ab}{(1 + b + ab)} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\frac{1}{1 + a + b^{-1}} + \frac{1}{1 + b + c^{-1}} + \frac{1}{1 + c + a^{-1}} = 1$.

Given,

$\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1$

Solving L.H.S :

$$\begin{aligned}\Rightarrow \sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} \\[1em] \Rightarrow \sqrt{\Big(\frac{1}{a}\Big)b} \times \sqrt{\Big(\frac{1}{b}\Big)c} \times \sqrt{\Big(\frac{1}{c}\Big)a} \\[1em] \Rightarrow \sqrt{\Big(\frac{b}{a}\Big)} \times \sqrt{\Big(\frac{c}{b}\Big)} \times \sqrt{\Big(\frac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\frac{b}{a}\Big) \times \Big(\frac{c}{b}\Big) \times \Big(\frac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{1} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence proved, $\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1$.

Given,

$\frac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3}= 3^{-3}$

Solving:

$$\begin{aligned}\Rightarrow \frac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3} = 3^{-3} \\[1em] \Rightarrow \frac{[(3)^2]^n \times 3^2 \times 3^n - [(3)^3]^n}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{(3)^{2n} \times 3^2 \times 3^n - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{(3)^{2n + 2 + n} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{(3)^{3n + 2} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{3^{3n}.3^2 - 3^{3n}}{3^{3m} \times 8} = 3^{-3}\\[1em] \Rightarrow \frac{3^{3n}[(3^2) - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{3^{3n}[9 - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \frac{3^{3n} \times 8}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow 3^{3n - 3m} = 3^{-3} \\[1em] \Rightarrow 3^{3(n - m)} = 3^{-3} \\[1em] \Rightarrow 3^{-3(m - n)} = 3^{-3} \\[1em]\end{aligned}$$

Equating the exponents,

⇒ -3(m - n) = -3

⇒ (m - n) = $\frac{-3}{-3}$ = 1.

Hence proved, m - n = 1.

Given,

21168 = x⁴ × y³ × z²

Factorizing 21168, we get :

⇒ 21168 = 2⁴ × 3³ × 7²

⇒ 2⁴ × 3³ × 7² = x⁴ × y³ × z²

⇒ x = 2, y = 3, z = 7

Hence, x = 2, y = 3, z = 7.

Given,

1960 = 2^a × 5^b × 7^c

Factorizing 1960, we get :

⇒ 1960 = 2³ × 5¹ × 7²

Equating the exponents,

⇒ 2³ × 5¹ × 7² = 2^a × 5^b × 7^c

⇒ a = 3, b = 1, c = 2.

The value of 2^-a × 5^-c × 7^b,

$$\begin{aligned}\Rightarrow \frac{1}{2^a} \times \frac{1}{5^c} \times 7^b \\[1em] \Rightarrow \frac{1}{2^3} \times \frac{1}{5^2} \times 7^1 \\[1em] \Rightarrow \frac{1}{8} \times \frac{1}{25} \times 7^1 \\[1em] \Rightarrow \frac{7}{200}.\end{aligned}$$

Hence, a = 3, b = 1, c = 2 and 2^-a × 5^-c × 7^b = $\frac{7}{200}$.

Solve for x:

Given,

$$\begin{aligned}\Rightarrow 3^x = \frac{1}{3} \\[1em] \Rightarrow 3^x = \frac{1}{3} \\[1em] \Rightarrow 3^x = 3^{-1}\end{aligned}$$

Equating the exponents,

⇒ x = -1.

Hence, x = -1.

Given,

⇒ 3^{(2x + 1)} = 1

⇒ 3^{(2x + 1)} = 1

⇒ 3^{(2x + 1)} = 3⁰

Equating the exponents,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\frac{1}{2}$

Hence, x = $-\frac{1}{2}$.

Given,

$$\begin{aligned}\Rightarrow \sqrt{\frac{a}{b}} = \Big(\frac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{\frac{1}{2}} = \Big(\frac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\frac{b}{a}\Big)^{-\frac{1}{2}} = \Big(\frac{b}{a}\Big)^{1 - 3x}\end{aligned}$$

Equating the exponents,

$$\begin{aligned}\Rightarrow -\frac{1}{2} = 1 - 3x \\[1em] \Rightarrow -1 = 2(1 - 3x) \\[1em] \Rightarrow -1 = 2 - 6x \\[1em] \Rightarrow 6x = 2 + 1 \\[1em] \Rightarrow 6x = 3 \\[1em] \Rightarrow x = \frac{3}{6} = \frac{1}{2}.\end{aligned}$$

Hence, x = $\frac{1}{2}$.

Given,

$$\begin{aligned}\Rightarrow \Big(\sqrt[3]{\frac{2}{3}}\Big)^{x-1} = \frac{27}{8} \\[1em] \Rightarrow \Big[\Big({\frac{2}{3}}\Big)^\frac{x - 1}{3}\Big] = \frac{3^3}{2^3} \\[1em] \Rightarrow \Big({\frac{2}{3}}\Big)^{\frac{x - 1}{3}} = \Big(\frac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big({\frac{2}{3}}\Big)^{\frac{x - 1}{3}} = \Big(\frac{2}{3}\Big)^{-3}\end{aligned}$$

Equating the exponents,

$$\begin{aligned}\Rightarrow \frac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.\end{aligned}$$

Hence, x = -8.

Given,

$$\begin{aligned}\Big(\sqrt\frac{3}{5}\Big)^{x-1} = \Big(\frac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big[\Big(\frac{3}{5}\Big)^{\frac{1}{2}}\Big]^{x-1} = \Big(\frac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big(\frac{3}{5}\Big)^{\frac{x - 1}{2}} = \Big(\frac{3^3}{5^3}\Big)^{-1} \\[1em] \Rightarrow \Big(\frac{3}{5}\Big)^{\frac{x - 1}{2}} = \Big(\frac{3}{5}\Big)^{3 \times (-1)} \\[1em] \Rightarrow \Big(\frac{3}{5}\Big)^{\frac{x - 1}{2}} = \Big(\frac{3}{5}\Big)^{-3}\end{aligned}$$

Equating the exponents,

$$\begin{aligned}\Rightarrow \frac{x - 1}{2} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 2 \\[1em] \Rightarrow x - 1 = -6 \\[1em] \Rightarrow x = -6 + 1 = -5.\end{aligned}$$

Hence, x = -5.

Given,

⇒ 9 × 3^x = (27)^{2x - 5}

⇒ 3² × 3^x = (3³)^{(2x - 5)}

⇒ 3^{(x + 2)} = 3^{(6x - 15)}

Equating the exponents,

⇒ x + 2 = 6x - 15

⇒ 6x - x = 15 + 2

⇒ 5x = 17

⇒ x = $\frac{17}{5} = 3\frac{2}{5}$.

Hence, x = $3\frac{2}{5}$.

Given,

⇒ 2^{(5x - 1)} = 4 × 2^{(3x + 1)}

⇒ 2^{(5x - 1)} = 4 × 2^{(3x + 1)}

⇒ 2^{(5x - 1)} = 2² × 2^{(3x + 1)}

⇒ 2^{(5x - 1)} = 2^{(3x + 1 + 2)}

⇒ 2^{(5x - 1)} = 2^{(3x + 3)}

Equating the exponents,

⇒ 5x - 1 = 3x + 3

⇒ 5x - 3x = 3 + 1

⇒ 2x = 4

⇒ x = $\frac{4}{2} = 2$

Hence, x = 2.

Given,

⇒ 5^{(x - 3)} × 3^{(2x - 8)} = 225

⇒ 5^{(x - 3)} × 3^{(2x - 8)} = 5² × 3²

Equating the exponents of 5,

⇒ x - 3 = 2

⇒ x = 2 + 3 = 5.

Equating the exponents of 3,

⇒ 2x - 8 = 2

⇒ 2x = 2 + 8

⇒ x = $\frac{10}{2}$ = 5.

Hence, x = 5.

Given,

$\frac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n - 1}} = 81$

Solving for n,

$$\begin{aligned}\Rightarrow \frac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n-1}} = 81 \\[1em] \Rightarrow \frac{3^n \times [(3)^2]^{n + 1}}{3^{n - 1} \times [(3)^3]^{n-1}} = 3^4 \\[1em] \Rightarrow \frac{3^n \times 3^{2n + 2}}{3^{n - 1} \times (3)^{3n - 3}} = 3^4 \\[1em] \Rightarrow \frac{3^{2n + 2 + n}}{3^{n - 1 + 3n - 3}} = 3^4 \\[1em] \Rightarrow \frac{3^{3n + 2}}{3^{4n - 4}} = 3^4 \\[1em] \Rightarrow 3^{3n + 2 - (4n - 4)}= 3^4 \\[1em] \Rightarrow 3^{3n + 2 - 4n + 4}= 3^4 \\[1em] \Rightarrow 3^{6 - n}= 3^4\end{aligned}$$

Equating the exponents,

⇒ 6 - n = 4

⇒ n = 6 - 4 = 2.

Hence, n = 2.

Given,

2^x = 3^y = 12^z

Let 2^x = 3^y = 12^z = k,

First term,

⇒ 2^x = k

⇒ x = 1

⇒ 2 = $k^{\frac{1}{x}}$

Second term,

⇒ 3^y = k

⇒ y = 1

⇒ 3 = $k^{\frac{1}{y}}$

Third term,

⇒ 12^z = k

⇒ z = 1

⇒ 12 = $k^{\frac{1}{z}}$

Factorizing 12, we get :

⇒ 12 = 2² × 3

We have $2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 12 = k^{\frac{1}{z}}$,

$$\begin{aligned}\Rightarrow k^{\frac{1}{z}} = \Big(k^{\frac{1}{x}}\Big)^2 + k^{\frac{1}{y}} \\[1em] \Rightarrow k^{\frac{1}{z}} = \Big(k^{\frac{2}{x}}\Big) + k^{\frac{1}{y}} \\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow \frac{1}{z} = \frac{2}{x} + \frac{1}{y}\end{aligned}$$

Hence proved, $\frac{1}{z} = \frac{1}{y} + \frac{2}{x}$.

Given,

2^x = 3^y = 6^-z

Let 2^x = 3^y = 6^-z = k,

First term,

⇒ 2^x = k

⇒ x = 1

⇒ 2 = $k^{\frac{1}{x}}$

Second term,

⇒ 3^y = k

⇒ y = 1

⇒ 3 = $k^{\frac{1}{y}}$

Third term,

⇒ 6^-z = k

⇒ z = -1

⇒ 6 = $k^{-\frac{1}{z}}$

Factorizing 6, we get :

⇒ 6 = 2 × 3

We have $2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 6 = k^{-\frac{1}{z}}$,

$$\begin{aligned}\Rightarrow k^{-\frac{1}{z}} = k^{\frac{1}{x}} \times k^{\frac{1}{y}} \\[1em] \Rightarrow k^{-\frac{1}{z}} = k^{\frac{1}{x}} \times k^{\frac{1}{y}}\\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow -\frac{1}{z} = \frac{1}{x} + \frac{1}{y} \\[1em] \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0.\end{aligned}$$

Hence proved, $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

Given,

$$\begin{aligned}\Rightarrow 128 \times 32^{\Big(\frac{-4}{3}\Big)} \\[1em] \Rightarrow (2)^7 \times [(2)^5]^{\frac{-4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\frac{5 \times -4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\frac{-20}{3}} \\[1em] \Rightarrow (2)^{7 - \frac{20}{3}} \\[1em] \Rightarrow (2)^{\frac{21 - 20}{3}} \\[1em] \Rightarrow (2)^{\frac{1}{3}} \\[1em] \Rightarrow \sqrt[3]{2}.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$\Big[\Big(\sqrt[3]{a^4}\Big)^{\frac{-3}{2}}\Big]^{\frac{-1}{2}}$

Simplifying the expression:

$$\begin{aligned}\Rightarrow \Big[\Big(a^4\Big)^{\frac{1}{3}}\Big]^{\frac{-3}{2} \times \frac{-1}{2}} \\[1em] \Rightarrow \Big[\Big(a^4\Big)^{\frac{1}{3}}\Big]^{\frac{3}{4}} \\[1em] \Rightarrow a^{4 \times \frac{1}{3} \times \frac{3}{4}} \\[1em] \Rightarrow a^1 \\[1em] \Rightarrow a.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

$\frac{\sqrt{5 \times 3^{-3}} \times \sqrt[6]{5 \times 3^6}}{\sqrt[6]{5} \times \sqrt{3}}$

Simplifying the expression:

$$\begin{aligned}\Rightarrow \frac{\sqrt{5} \times \sqrt{3^{-3}} \times \sqrt[6]{5} \times \sqrt[6]{3^6}}{\sqrt[6]{5} \times \sqrt{3}} \\[1em] \Rightarrow \frac{\sqrt{5} \times {(3^{-3})}^{\frac{1}{2}} \times {3}^{6 \times \frac{1}{6}}}{\sqrt{3}} \\[1em] \Rightarrow \frac{\sqrt{5} \times {3}^{\frac{-3}{2}} \times {3^1}}{{3}^{\frac{1}{2}}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\frac{-3}{2} + 1 - \frac{1}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\frac{-3 + 2 - 1}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\frac{-2}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{-1} \\[1em] \Rightarrow \frac {\sqrt{5}}{3}.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

⇒ (81)^0.13 × (81)^0.12

Simplifying the expression:

⇒ (81)^{(0.13 + 0.12)}

⇒ (81)^0.25

⇒ [(3)⁴]^0.25

⇒ 3^{4 × 0.25}

⇒ 3¹

⇒ 3.

Hence, option 2 is the correct option.

Given,

⇒ 3^x = 3^(-x)

$$\begin{aligned}\Rightarrow 3^x = \frac{1}{3^x} \\[1em] \Rightarrow 3^x \times 3^x = 1 \\[1em] \Rightarrow 3^{2x} = 1 \\[1em] \Rightarrow 3^{2x} = 3^0 \\[1em]\end{aligned}$$

Equating the exponents,

$$\begin{aligned}\Rightarrow 2x = 0 \\[1em] \Rightarrow x = 0\end{aligned}$$

Substituting value of x in (1.2)^x, we get :

⇒ (1.2)⁰

⇒ 1.

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow 9 \times 81^x = \frac{1}{27^{x - 3}} \\[1em] \Rightarrow 3^2 \times (3^4)^{x} = \frac{1}{3^{3(x - 3)}} \\[1em] \Rightarrow 3^2 \times 3^{4x} = 3^{-3(x - 3)} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3 \times x - 3 \times - 3} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3x + 9}\end{aligned}$$

Equating the exponents:

$$\begin{aligned}\Rightarrow 2 + 4x = -3x + 9 \\[1em] \Rightarrow 4x + 3x = 9 - 2 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \frac{7}{7} \\[1em] \Rightarrow x = 1.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

⇒ 4 × 2^{x + 3} = 8^{x + 1}

Simplifying the expression,

$$\begin{aligned}\Rightarrow 4 \times 2^{x + 3} = (2^3)^{x + 1} \\[1em] \Rightarrow 4 \times 2^{x + 3} = 2^{3x + 3} \\[1em] \Rightarrow 4 = \frac{2^{3x + 3}}{2^{x + 3} } \\[1em] \Rightarrow 2^2 = 2^{3x + 3} \times 2^{-(x + 3)} \\[1em] \Rightarrow 2^2 = 2^{3x + 3 - (x + 3)} \\[1em] \Rightarrow 2^2 = 2^{3x - x + 3 - 3} \\[1em] \Rightarrow 2^2 = 2^{2x}\end{aligned}$$

Equating the exponents:

⇒ 2 = 2x

⇒ x = $\frac{2}{2}$

⇒ x = 1.

⇒ 2^x = 2¹ = 2.

Hence, option 2 is the correct option.

Given,

2^{(x + 3)} + 2^{(x + 1)} = 320

Simplifying the expression:

$$\begin{aligned}\Rightarrow 2^{x + 1 + 2} + 2^{x + 1} = 320 \\[1em] \Rightarrow 2^{x + 1} \times 2^2 + 2^{x + 1} = 320 \\[1em] \Rightarrow 2^{x + 1} \Big(2^2 + 1\Big) = 320 \\[1em] \Rightarrow 2^{x + 1} \times 5 = 320 \\[1em] \Rightarrow 2^{x + 1} = \frac{320}{5} \\[1em] \Rightarrow 2^{x + 1} = 64 \\[1em] \Rightarrow 2^{x + 1} = 2^6\end{aligned}$$

Equating the exponents:

⇒ x + 1 = 6

⇒ x = 6 - 1

⇒ x = 5.

Hence, option 4 is the correct option.

Given,

⇒ 4^x = 8^y

Simplifying the expression:

⇒ (2²)^x = (2³)^y

⇒ 2^2x = 2^3y

Equating the exponents:

$$\begin{aligned}\Rightarrow 2x = 3y \\[1em] \Rightarrow \frac{x}{y} = \frac{3}{2} \\[1em] \Rightarrow x : y = 3 : 2.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

(2⁵ + 0.125)² - (2⁵ - 0.125)² = 2^x

By using the identity:

(a + b)² - (a - b)² = 4ab

Let a = 2⁵, b = 0.125

Using identity in L.H.S. of the given equation,

$$\begin{aligned}\Rightarrow (2^5 + 0.125)^2 - (2^5 - 0.125)^2 = 4 \times 2^5 \times 0.125 \\[1em] = 4 \times 32 \times \frac{125}{1000} \\[1em] = 4 \times 32 \times \frac{1}{8} \\[1em] = 16 \\[1em] = 2^4.\end{aligned}$$

Equation L.H.S. and R.H.S.,

⇒ 2^x = 2⁴

⇒ x = 4.

Hence, option 3 is the correct option.

Given,

2^{x + 1} + 2^x = 3

Now simplifying:

⇒ 2 × 2^x + 2^x = 3

⇒ 2^x(2 + 1) = 3

⇒ 2^x × 3 = 3

⇒ 2^x = 1

⇒ 2^x = 2⁰

Equating the exponents:

⇒ x = 0

Substituting value of x in 3^x + 3^-x, we get :

⇒ 3⁰ + 3⁰

⇒ 1 + 1

⇒ 2.

Hence, option 3 is the correct option.

Let us consider l^x = m^y = n^z = k, and lmn = 1

From, l^x = k, we get $l = k^{\frac{1}{x}}$

From, m^y = k, we get $m = k^{\frac{1}{y}}$

From, n^z = k, we get $n = k^{\frac{1}{z}}$

Substituting in lmn = 1:

$$\begin{aligned}\Rightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}} \times k^{\frac{1}{z}} = 1 \\[1em] \Rightarrow k^{\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)} = 1 \\[1em] \Rightarrow k^{\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)} = k^0\end{aligned}$$

Equating the exponents:

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Multiply both sides by xyz:

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \\[1em] \Rightarrow xyz \Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\Big) = 0 \times xyz \\[1em] \Rightarrow \Big(\frac{xyz}{x} + \frac{xyz}{y} + \frac{xyz}{z}\Big) = 0 \\[1em] \Rightarrow yz + zx + xy = 0.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

$\frac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x}$

Solving Numerator:

⇒ 9 × (4^x)²

⇒ 9 × [(2²)^x]²

⇒ 9 × 2^4x

Solving Denominator:

⇒ 16^{x + 1} - 2^{x + 1} × 8^x

⇒ (2⁴)^{x + 1} - 2^{x + 1} × (2³)^x

⇒ 2^{4x + 4} - 2^{x + 1} × 2^3x

⇒ 2^{4x + 4} - 2^{x + 1 + 3x}

⇒ 2^{4x + 4} - 2^{4x + 1}

Substituting the simplified numerator and denominator in the original expression:

$$\begin{aligned}\Rightarrow \frac{9 \times 2^{4x}}{2^{4x + 4} - 2^{4x + 1}} \\[1em] \Rightarrow \frac{9 \times 2^{4x}}{2^{4x}.2^4 - 2^{4x}.2^1} \\[1em] \Rightarrow \frac{9 \times 2^{4x}}{2^{4x}(2^4 - 2^1)} \\[1em] \Rightarrow \frac{9 \times 2^{4x}}{2^{4x} \times (16 - 2)} \\[1em] \Rightarrow \frac{9}{14}.\end{aligned}$$

Hence, option 4 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow [1 - (1 - [1 - x^3]^{(-1)})^{(-1)}]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(1 - \frac{1}{1 - x^3}\Big)^{-1}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\frac{1 - x^3 - 1}{1 - x^3}\Big)^{-1}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\frac{-x^3}{1 - x^3}\Big)^{-1}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\frac{x^3}{x^3 - 1}\Big)^{-1}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[1 - \frac{x^3 - 1}{x^3}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[\frac{x^3 - x^3 + 1}{x^3}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow \Big[\frac{1}{x^3}\Big]^{-\frac{1}{3}} \\[1em] \Rightarrow (x^3)^{\frac{1}{3}} \\[1em] \Rightarrow x \\[1em] \Rightarrow 0.1\end{aligned}$$

Hence, option 3 is the correct option.

Given,

⇒ (x^a)^{(b - c)}(x^b)^{(c - a)}(x^c)^{(a - b)}

⇒ x^{(ab - ac)}x^{(bc - ba)}x^{(ca - cb)}

⇒ x^{(ab - ac) + (bc - ba) + (ca - cb)}

⇒ x^{(ab - ac + bc - ba + ca - cb)}

⇒ x^{(ab - ab + bc -bc - ac + ac)}

⇒ x⁰

⇒ 1.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{8}{162}\Big)^{-1.5} \\[1em] \Rightarrow \Big(\frac{2^3}{2 \times 9^2}\Big)^{-1.5} \\[1em] \Rightarrow \Big(\frac{2^{3 - 1}}{9^2}\Big)^{-1.5} \\[1em] \Rightarrow \Big(\frac{2^2}{9^2}\Big)^{-1.5} \\[1em] \Rightarrow \Big[\Big(\frac{2}{9}\Big)^2\Big]^{-\frac{3}{2}} \\[1em] \Rightarrow \Big(\frac{2}{9}\Big)^{2 \times -\frac{3}{2}} \\[1em] \Rightarrow \Big(\frac{2}{9}\Big)^{-3} \\[1em] \Rightarrow \Big(\frac{9}{2}\Big)^{3} \\[1em] \Rightarrow \frac{729}{8}.\end{aligned}$$

Assertion (A) is true.

x^m × yⁿ = (xy)^{m + n} is not correct in general.

Correct identity:

⇒ x^m × y^m = (xy)^m

Or

⇒ x^m × xⁿ = x^m+n

Reason (R) is false.

A is true, R is false

Hence, option 1 is the correct option.

Given,

(x^-1 - y^-1) × (x - y)^-1 = x^-1y^-1

Solving L.H.S,

$$\begin{aligned}\Rightarrow (x^{-1} - y^{-1}) \times (x - y)^{-1} \\[1em] \Rightarrow \Big(\frac{1}{x} - \frac{1}{y}\Big) \times \Big(\frac{1}{x - y}\Big) \\[1em] \Rightarrow \Big(\frac{y - x}{xy}\Big) \times \frac{1}{x - y} \\[1em] \Rightarrow \frac{-(x - y)}{xy} \times \frac{1}{x - y} \\[1em] \Rightarrow \frac{-1}{xy} \\[1em] \Rightarrow -x^{-1}y^{-1}.\end{aligned}$$

L.H.S ≠ R.H.S

Thus, Assertion (A) is false.

For any non-zero number x, the notation x⁻¹ is defined as its multiplicative inverse, which is equal to $\frac{1}{x}$

Thus, Reason (R) is true.

A is false, R is true

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{3^{-2}}{3^4} \\[1em] \Rightarrow 3^{-2 - 4} \\[1em] \Rightarrow 3^{-6} \\[1em] \Rightarrow (3^2)^{-3} \\[1em] \Rightarrow 9^{-3}.\end{aligned}$$

∴ Assertion (A) is false.

According to reason (R) :

$\Big(\frac{x}{y}\Big)^{-m} = \frac{y^{-m}}{x^{-m}}$

Solving L.H.S,

$$\begin{aligned}\Rightarrow \Big(\frac{x}{y}\Big)^{-m} \\[1em] \Rightarrow \Big(\frac{y}{x}\Big)^{m} \\[1em] \Rightarrow \Big(\frac{y^m}{x^m}\Big) \\[1em]\end{aligned}$$

Since, L.H.S ≠ R.H.S

∴ Reason (R) is false.

Both A and R are false.

Hence, option 4 is the correct option.

Solving Option 2,

Given,

$\Big(\sqrt{x^3}\Big)^{\frac{2}{3}}$

Now simplifying:

$$\begin{aligned}\Rightarrow ({x^3})^{\frac{1}{2} \times \frac{2}{3}} \\[1em] \Rightarrow ({x^3})^{\frac{1}{3}} \\[1em] \Rightarrow (x)^{3 \times \frac{1}{3}} \\[1em] \Rightarrow x.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$\Big[5\Big(8^\frac{1}{3} + 27^\frac{1}{3}\Big)^{3}\Big]^{\frac{1}{4}}$

Now simplifying:

$$\begin{aligned}\Rightarrow \Big[5\Big((2^3)^{\frac{1}{3}} + (3^3)^{\frac{1}{3}}\Big)^{3}\Big]^{\frac{1}{4}} \\[1em] \Rightarrow \Big[5(2 + 3)^{3}\Big]^{\frac{1}{4}} \\[1em] \Rightarrow \Big[5(5)^{3}\Big]^{\frac{1}{4}} \\[1em] \Rightarrow \Big[(5)^{4}\Big]^{\frac{1}{4}} \\[1em] \Rightarrow 5^{4 \times \frac{1}{4}} \\[1em] \Rightarrow 5.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

4.2 × 10^-15 + 42 × 10^-16 + 4.2 × 10^-14

Now simplifying:

$$\begin{aligned}\Rightarrow 10^{-14} (4.2 × 10^{-1} + 42 × 10^{-2} + 4.2) \\[1em] \Rightarrow 10^{-14} (4.2 \times \frac{1}{10} + 42 \times \frac{1}{100} + 4.2) \\[1em] \Rightarrow 10^{-14} \Big(\frac{4.2}{10} + \frac{42}{100} + 4.2\Big) \\[1em] \Rightarrow 10^{-14} (0.42 + 0.42 + 4.2) \\[1em] \Rightarrow 5.04 × 10^{-14}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

⇒ (2^-1 × 2² × 2^-3 × 2⁴ × 2^-5 × 2⁶ × 2^-7 × 2⁸ × 2^-9 × 2¹⁰)^-1

⇒ [2^{(-1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 - 9 + 10)}]^-1

⇒ [2^{(-1 - 3 - 5 - 7 - 9 + 2 + 4 + 6 + 8 + 10)}]^-1

⇒ [2^{(-25 + 30)}]^-1

⇒ (2⁵)^-1

⇒ 2^-5

⇒ $\frac{1}{2^5}$

⇒ $\frac{1}{32}$.

Hence, option 3 is the correct option.

Given,

(-1)⁰ - (-1)¹ - (-1)² - (-1)³ - (-1)⁴ - (-1)⁵ - (-1)⁶ - (-1)⁷ - (-1)⁸ - (-1)⁹ - (-1)¹⁰

A negative number raised to an even power gives a positive result, also a real number raised to power 0 results in 1.

A negative number raised to an odd power gives a negative result.

Now simplifying:

⇒ 1 - (-1) - (1) - (-1) - (1) - (-1) - (1) - (-1) - (1) - (-1) - (1)

⇒ 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1

⇒ 1.

Hence, option 2 is the correct option.

Simplifying option 3,

⇒ 15² × 15^-2

⇒ 15^{2 - 2}

⇒ 15⁰

⇒ 1.

Hence, option 3 is the correct option.

Yes, if the base is a fraction between 0 and 1, raising it to a negative power results in a large number.

Example 1: Let the number be 0.2

Then, 0.2^-1

⇒ $\frac{1}{0.2}$

⇒ 5, which is greater than original number.

Hence, yes if the base is a fraction between 0 and 1.

Given,

$\Rightarrow (x - 1)^{\frac{1}{2}} + (y - 2)^{\frac{1}{2}} + (z - 3)^{\frac{1}{2}} = 0$

We can write,

⇒ $(x - 1)^{\frac{1}{2}}$= 0 ....(1)

⇒ $(y - 2)^{\frac{1}{2}}$ = 0 ....(2)

⇒ $(z - 3)^{\frac{1}{2}}$ = 0 ....(3)

Solving equation 1,

⇒ $(x - 1)^{\frac{1}{2}}$= 0

Squaring both sides,

⇒ (x - 1) = 0

⇒ x = 1

Solving equation 2,

⇒ $(y - 2)^{\frac{1}{2}}$ = 0

Squaring both sides,

⇒ (y - 2) = 0

⇒ y = 2

Solving equation 3,

⇒ $(z - 3)^{\frac{1}{2}}$ = 0

Squaring both sides,

⇒ (z - 3) = 0

⇒ z = 3

Hence, the values of x, y, z are 1, 2, 3.