Cubes and Cube Roots

Finding prime factors of 648 :

$$\begin{array}{l|r} 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

648 = (2 × 2 × 2) × (3 × 3 × 3) × 3

Since 3 is left after grouping the prime factors in triplets, so 648 cannot be expressed as the product of triplets of equal prime factors.

Hence, 648 is not a perfect cube.

Finding prime factors of 729 :

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

729 = (3 × 3 × 3) × (3 × 3 × 3)

Since no prime factor is left after grouping the prime factors in triplets, so 729 can be expressed as the product of triplets of equal prime factors.

Hence, 729 is a perfect cube.

Finding prime factors of 8640 :

$$\begin{array}{l|r} 2 & 8640 \\ \hline 2 & 4320 \\ \hline 2 & 2160 \\ \hline 2 & 1080 \\ \hline 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

8640 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5

Since 5 is left after grouping the prime factors in triplets, so 8640 cannot be expressed as the product of triplets of equal prime factors.

Hence, 8640 is not a perfect cube.

Finding prime factors of 8000 :

$$\begin{array}{l|r} 2 & 8000 \\ \hline 2 & 4000 \\ \hline 2 & 2000 \\ \hline 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

8000 = (2 × 2 × 2) × (2 × 2 × 2) × (5 × 5 × 5)

Since no prime factor is left after grouping the prime factors in triplets, so 8000 can be expressed as the product of triplets of equal prime factors.

Hence, 8000 is a perfect cube.

Finding prime factors of 1728 :

$$\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Since 1728 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{1728}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3\\[1em] \Rightarrow 12\end{aligned}$$

Hence, 1728 is a perfect cube and $\mathbf{\sqrt[3]{1728} = 12}$.

Finding prime factors of 5832 :

$$\begin{array}{l|r} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

Since 5832 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{5832}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 3 \times 3\\[1em] \Rightarrow 18\end{aligned}$$

Hence, 5832 is a perfect cube and $\mathbf{\sqrt[3]{5832} = 18}$.

Finding prime factors of 13824 :

$$\begin{array}{l|r} 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

13824 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Since 13824 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{13824}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow 24\end{aligned}$$

Hence, 13824 is a perfect cube and $\mathbf{\sqrt[3]{13824} = 24}$.

Finding prime factors of 35937 :

$$\begin{array}{l|r} 3 & 35937 \\ \hline 3 & 11979 \\ \hline 3 & 3993 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

35937 = (3 × 3 × 3) × (11 × 11 × 11)

Since 35937 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{35937}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (11 \times 11 \times 11)}\\[1em] \Rightarrow 3 \times 11\\[1em] \Rightarrow 33\end{aligned}$$

Hence, 35937 is a perfect cube and $\mathbf{\sqrt[3]{35937} = 33}$.

Finding prime factors of 243 :

$$\begin{array}{l|r} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

243 = (3 × 3 × 3) × 3 × 3

Since 3 × 3 is left after grouping the prime factors in triplets, so one more 3 is needed to complete the triplet.

243 × 3 = 729 = (3 × 3 × 3) × (3 × 3 × 3), which is a perfect cube.

Hence, the smallest number by which 243 must be multiplied to obtain a perfect cube is 3.

Finding prime factors of 3072 :

$$\begin{array}{l|r} 2 & 3072 \\ \hline 2 & 1536 \\ \hline 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

3072 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2 × 3

Since 2 and 3 are left after grouping the prime factors in triplets, so two more 2's and two more 3's are needed to complete the triplets.

∴ The required number = 2 × 2 × 3 × 3 = 36

3072 × 36 = 110592 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3), which is a perfect cube.

Hence, the smallest number by which 3072 must be multiplied to obtain a perfect cube is 36.

Finding prime factors of 11979 :

$$\begin{array}{l|r} 3 & 11979 \\ \hline 3 & 3993 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

11979 = 3 × 3 × (11 × 11 × 11)

Since 3 × 3 is left after grouping the prime factors in triplets, so one more 3 is needed to complete the triplet.

11979 × 3 = 35937 = (3 × 3 × 3) × (11 × 11 × 11), which is a perfect cube.

Hence, the smallest number by which 11979 must be multiplied to obtain a perfect cube is 3.

Finding prime factors of 19652 :

$$\begin{array}{l|r} 2 & 19652 \\ \hline 2 & 9826 \\ \hline 17 & 4913 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$$

19652 = 2 × 2 × (17 × 17 × 17)

Since 2 × 2 is left after grouping the prime factors in triplets, so one more 2 is needed to complete the triplet.

19652 × 2 = 39304 = (2 × 2 × 2) × (17 × 17 × 17), which is a perfect cube.

Hence, the smallest number by which 19652 must be multiplied to obtain a perfect cube is 2.

Finding prime factors of 1536 :

$$\begin{array}{l|r} 2 & 1536 \\ \hline 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3

Since 3 is left after grouping the prime factors in triplets, so 1536 should be divided by 3.

1536 ÷ 3 = 512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.

Finding prime factors of 10985 :

$$\begin{array}{l|r} 5 & 10985 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

10985 = 5 × (13 × 13 × 13)

Since 5 is left after grouping the prime factors in triplets, so 10985 should be divided by 5.

10985 ÷ 5 = 2197 = (13 × 13 × 13), which is a perfect cube.

Hence, the smallest number by which 10985 must be divided to obtain a perfect cube is 5.

Finding prime factors of 28672 :

$$\begin{array}{l|r} 2 & 28672 \\ \hline 2 & 14336 \\ \hline 2 & 7168 \\ \hline 2 & 3584 \\ \hline 2 & 1792 \\ \hline 2 & 896 \\ \hline 2 & 448 \\ \hline 2 & 224 \\ \hline 2 & 112 \\ \hline 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

28672 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 7

Since 7 is left after grouping the prime factors in triplets, so 28672 should be divided by 7.

28672 ÷ 7 = 4096 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 28672 must be divided to obtain a perfect cube is 7.

Finding prime factors of 13718 :

$$\begin{array}{l|r} 2 & 13718 \\ \hline 19 & 6859 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array}$$

13718 = 2 × (19 × 19 × 19)

Since 2 is left after grouping the prime factors in triplets, so 13718 should be divided by 2.

13718 ÷ 2 = 6859 = (19 × 19 × 19), which is a perfect cube.

Hence, the smallest number by which 13718 must be divided to obtain a perfect cube is 2.

Volume of one cuboid = 3 × 3 × 5 = 45 cm³

Expressing 45 into prime factors, we have

45 = 3 × 3 × 5

To form a cube, the prime factors must occur in triplets.

Here, one more 3 and two more 5's are needed to complete the triplets.

∴ Number of cuboids needed = 3 × 5 × 5 = 75

Volume of the cube so formed = 45 × 75 = 3375 = 15 × 15 × 15 cm³

Hence, 75 such cuboids will be needed to form a cube.

Let the length of the side of the cubical box be $a$ cm.

Surface area of the cubical box = $6a^2$

$$\begin{aligned}\Rightarrow 6a^2 = 486\\[1em] \Rightarrow a^2 = \frac{486}{6}\\[1em] \Rightarrow a^2 = 81\\[1em] \Rightarrow a = \sqrt{81}\\[1em] \Rightarrow a = 9\end{aligned}$$

Volume of the cubical box = $a^3$ = 9 × 9 × 9 = 729

Hence, the volume of the cubical box is 729 cm³.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

125 = 5 × 5 × 5 = (5)³

Since 5 is an odd natural number,

Hence, 125 is the cube of the odd natural number 5.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

512 = 8 × 8 × 8 = (8)³

Since 8 is an even natural number,

Hence, 512 is the cube of the even natural number 8.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

1000 = 10 × 10 × 10 = (10)³

Since 10 is an even natural number,

Hence, 1000 is the cube of the even natural number 10.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

2197 = 13 × 13 × 13 = (13)³

Since 13 is an odd natural number,

Hence, 2197 is the cube of the odd natural number 13.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

4096 = 16 × 16 × 16 = (16)³

Since 16 is an even natural number,

Hence, 4096 is the cube of the even natural number 16.

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

6859 = 19 × 19 × 19 = (19)³

Since 19 is an odd natural number,

Hence, 6859 is the cube of the odd natural number 19.

If a number has 1 in the unit place, then its cube also ends in 1.

One's digit of 231 = 1

Cube of one's digit = 1 × 1 × 1 = 1

Hence, the one's digit of the cube of 231 is 1.

If a number has 8 in the unit place, then its cube ends in 2.

One's digit of 358 = 8

Cube of one's digit = 8 × 8 × 8 = 512

Hence, the one's digit of the cube of 358 is 2.

If a number has 9 in the unit place, then its cube also ends in 9.

One's digit of 419 = 9

Cube of one's digit = 9 × 9 × 9 = 729

Hence, the one's digit of the cube of 419 is 9.

If a number has 5 in the unit place, then its cube also ends in 5.

One's digit of 725 = 5

Cube of one's digit = 5 × 5 × 5 = 125

Hence, the one's digit of the cube of 725 is 5.

If a number has 4 in the unit place, then its cube also ends in 4.

One's digit of 854 = 4

Cube of one's digit = 4 × 4 × 4 = 64

Hence, the one's digit of the cube of 854 is 4.

If a number has 7 in the unit place, then its cube ends in 3.

One's digit of 987 = 7

Cube of one's digit = 7 × 7 × 7 = 343

Hence, the one's digit of the cube of 987 is 3.

If a number has 2 in the unit place, then its cube ends in 8.

One's digit of 752 = 2

Cube of one's digit = 2 × 2 × 2 = 8

Hence, the one's digit of the cube of 752 is 8.

If a number has 3 in the unit place, then its cube ends in 7.

One's digit of 893 = 3

Cube of one's digit = 3 × 3 × 3 = 27

Hence, the one's digit of the cube of 893 is 7.

Solving,

$$\begin{aligned}\Rightarrow (-13)^3\\[1em] \Rightarrow (-13) \times (-13) \times (-13)\\[1em] \Rightarrow -2197\end{aligned}$$

Hence, $\mathbf{(-13)^3 = -2197}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(3\frac{1}{5}\Big)^3\\[1em] \Rightarrow \Big(\frac{16}{5}\Big)^3\\[1em] \Rightarrow \frac{16 \times 16 \times 16}{5 \times 5 \times 5}\\[1em] \Rightarrow \frac{4096}{125}\\[1em] \Rightarrow 32\frac{96}{125}\end{aligned}$$

Hence, $\mathbf{\Big(3\frac{1}{5}\Big)^3 = 32\frac{96}{125}}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(-5\frac{1}{7}\Big)^3\\[1em] \Rightarrow \Big(\frac{-36}{7}\Big)^3\\[1em] \Rightarrow \frac{(-36) \times (-36) \times (-36)}{7 \times 7 \times 7}\\[1em] \Rightarrow \frac{-46656}{343}\\[1em] \Rightarrow -136\frac{8}{343}\end{aligned}$$

Hence, $\mathbf{\Big(-5\frac{1}{7}\Big)^3 = -136\frac{8}{343}}$.

Given number is 12167.

Finding prime factors of 12167 :

$$\begin{array}{l|r} 23 & 12167 \\ \hline 23 & 529 \\ \hline 23 & 23 \\ \hline & 1 \\ \end{array}$$

Expressing 12167 into prime factors, we have

12167 = 23 × 23 × 23

$$\begin{aligned}\Rightarrow \sqrt[3]{12167}\\[1em] \Rightarrow \sqrt[3]{23 \times 23 \times 23}\\[1em] \Rightarrow 23\end{aligned}$$

Hence, cube root of 12167 = 23.

Given number is 35937.

Finding prime factors of 35937 :

$$\begin{array}{l|r} 3 & 35937 \\ \hline 3 & 11979 \\ \hline 3 & 3993 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Expressing 35937 into prime factors, we have

35937 = (3 × 3 × 3) × (11 × 11 × 11)

$$\begin{aligned}\Rightarrow \sqrt[3]{35937}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (11 \times 11 \times 11)}\\[1em] \Rightarrow 3 \times 11\\[1em] \Rightarrow 33\end{aligned}$$

Hence, cube root of 35937 = 33.

Given number is 42875.

Finding prime factors of 42875 :

$$\begin{array}{l|r} 5 & 42875 \\ \hline 5 & 8575 \\ \hline 5 & 1715 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 42875 into prime factors, we have

42875 = (5 × 5 × 5) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{42875}\\[1em] \Rightarrow \sqrt[3]{(5 \times 5 \times 5) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 5 \times 7\\[1em] \Rightarrow 35\end{aligned}$$

Hence, cube root of 42875 = 35.

Given number is 21952.

Finding prime factors of 21952 :

$$\begin{array}{l|r} 2 & 21952 \\ \hline 2 & 10976 \\ \hline 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 21952 into prime factors, we have

21952 = (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{21952}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 7\\[1em] \Rightarrow 28\end{aligned}$$

Hence, cube root of 21952 = 28.

Given number is 373248.

Finding prime factors of 373248 :

$$\begin{array}{l|r} 2 & 373248 \\ \hline 2 & 186624 \\ \hline 2 & 93312 \\ \hline 2 & 46656 \\ \hline 2 & 23328 \\ \hline 2 & 11664 \\ \hline 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Expressing 373248 into prime factors, we have

373248 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

$$\begin{aligned}\Rightarrow \sqrt[3]{373248}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow 72\end{aligned}$$

Hence, cube root of 373248 = 72.

Given number is 32768.

Finding prime factors of 32768 :

$$\begin{array}{l|r} 2 & 32768 \\ \hline 2 & 16384 \\ \hline 2 & 8192 \\ \hline 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array}$$

Expressing 32768 into prime factors, we have

32768 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

$$\begin{aligned}\Rightarrow \sqrt[3]{32768}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] \Rightarrow 2 \times 2 \times 2 \times 2 \times 2\\[1em] \Rightarrow 32\end{aligned}$$

Hence, cube root of 32768 = 32.

Given number is 262144.

Finding prime factors of 262144 :

$$\begin{array}{l|r} 2 & 262144 \\ \hline 2 & 131072 \\ \hline 2 & 65536 \\ \hline 2 & 32768 \\ \hline 2 & 16384 \\ \hline 2 & 8192 \\ \hline 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array}$$

Expressing 262144 into prime factors, we have

262144 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

$$\begin{aligned}\Rightarrow \sqrt[3]{262144}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] \Rightarrow 2 \times 2 \times 2 \times 2 \times 2 \times 2\\[1em] \Rightarrow 64\end{aligned}$$

Hence, cube root of 262144 = 64.

Given number is 157464.

Finding prime factors of 157464 :

$$\begin{array}{l|r} 2 & 157464 \\ \hline 2 & 78732 \\ \hline 2 & 39366 \\ \hline 3 & 19683 \\ \hline 3 & 6561 \\ \hline 3 & 2187 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Expressing 157464 into prime factors, we have

157464 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3) × (3 × 3 × 3)

$$\begin{aligned}\Rightarrow \sqrt[3]{157464}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 3 \times 3 \times 3\\[1em] \Rightarrow 54\end{aligned}$$

Hence, cube root of 157464 = 54.

Given number is -250047.

Finding prime factors of 250047 :

$$\begin{array}{l|r} 3 & 250047 \\ \hline 3 & 83349 \\ \hline 3 & 27783 \\ \hline 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 250047 into prime factors, we have

250047 = (3 × 3 × 3) × (3 × 3 × 3) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{-250047}\\[1em] \Rightarrow -\sqrt[3]{250047}\\[1em] \Rightarrow -\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow -(3 \times 3 \times 7)\\[1em] \Rightarrow -63\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{-250047} = -63}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{\frac{-64}{1331}}\\[1em] \Rightarrow \frac{\sqrt[3]{-64}}{\sqrt[3]{1331}}\\[1em] \Rightarrow \frac{-\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{11 \times 11 \times 11}}\\[1em] \Rightarrow \frac{-(2 \times 2)}{11}\\[1em] \Rightarrow \frac{-4}{11}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{\frac{-64}{1331}} = \frac{-4}{11}}$.

Converting mixed fraction to improper fraction:

$4\frac{17}{27} = \frac{125}{27}$

$$\begin{aligned}\Rightarrow \sqrt[3]{4\frac{17}{27}}\\[1em] \Rightarrow \sqrt[3]{\frac{125}{27}}\\[1em] \Rightarrow \frac{\sqrt[3]{125}}{\sqrt[3]{27}}\\[1em] \Rightarrow \frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{3 \times 3 \times 3}}\\[1em] \Rightarrow \frac{5}{3}\\[1em] \Rightarrow 1\frac{2}{3}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{4\frac{17}{27}} = 1\frac{2}{3}}$.

Converting mixed fraction to improper fraction:

$5\frac{1182}{2197} = \frac{12167}{2197}$

$$\begin{aligned}\Rightarrow \sqrt[3]{5\frac{1182}{2197}}\\[1em] \Rightarrow \sqrt[3]{\frac{12167}{2197}}\\[1em] \Rightarrow \frac{\sqrt[3]{12167}}{\sqrt[3]{2197}}\\[1em] \Rightarrow \frac{\sqrt[3]{23 \times 23 \times 23}}{\sqrt[3]{13 \times 13 \times 13}}\\[1em] \Rightarrow \frac{23}{13}\\[1em] \Rightarrow 1\frac{10}{13}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{5\frac{1182}{2197}} = 1\frac{10}{13}}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt[3]{512 \times 729}\\[1em] \Rightarrow \sqrt[3]{512} \times \sqrt[3]{729}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)} \times \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow (2 \times 2 \times 2) \times (3 \times 3)\\[1em] \Rightarrow 8 \times 9\\[1em] \Rightarrow 72\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{512 \times 729} = 72}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt[3]{(-1331) \times (3375)}\\[1em] \Rightarrow \sqrt[3]{-1331} \times \sqrt[3]{3375}\\[1em] \Rightarrow -\sqrt[3]{11 \times 11 \times 11} \times \sqrt[3]{(3 \times 3 \times 3) \times (5 \times 5 \times 5)}\\[1em] \Rightarrow (-11) \times (3 \times 5)\\[1em] \Rightarrow (-11) \times 15\\[1em] \Rightarrow -165\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{(-1331) \times (3375)} = -165}$.

Converting decimal to fraction:

$0.003375 = \frac{3375}{1000000}$

$$\begin{aligned}\Rightarrow \sqrt[3]{0.003375}\\[1em] \Rightarrow \sqrt[3]{\frac{3375}{1000000}}\\[1em] \Rightarrow \frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}\\[1em] \Rightarrow \frac{\sqrt[3]{(3 \times 3 \times 3) \times (5 \times 5 \times 5)}}{\sqrt[3]{100 \times 100 \times 100}}\\[1em] \Rightarrow \frac{3 \times 5}{100}\\[1em] \Rightarrow \frac{15}{100}\\[1em] \Rightarrow 0.15\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{0.003375} = 0.15}$.

Converting decimal to fraction:

$19.683 = \frac{19683}{1000}$

$$\begin{aligned}\Rightarrow \sqrt[3]{19.683}\\[1em] \Rightarrow \sqrt[3]{\frac{19683}{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{19683}}{\sqrt[3]{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow \frac{3 \times 3 \times 3}{10}\\[1em] \Rightarrow \frac{27}{10}\\[1em] \Rightarrow 2.7\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{19.683} = 2.7}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}\\[1em] \Rightarrow \sqrt[3]{3 \times 3 \times 3} + \sqrt[3]{\frac{8}{1000}} + \sqrt[3]{\frac{64}{1000}}\\[1em] \Rightarrow 3 + \frac{\sqrt[3]{2 \times 2 \times 2}}{\sqrt[3]{10 \times 10 \times 10}} + \frac{\sqrt[3]{4 \times 4 \times 4}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow 3 + \frac{2}{10} + \frac{4}{10}\\[1em] \Rightarrow 3 + 0.2 + 0.4\\[1em] \Rightarrow 3.6\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} = 3.6}$.

Finding prime factors of 6561 :

$$\begin{array}{l|r} 3 & 6561 \\ \hline 3 & 2187 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

6561 = (3 × 3 × 3) × (3 × 3 × 3) × 3 × 3

Since 3 × 3 is left after grouping the prime factors in triplets, so one more 3 is needed to complete the triplet.

∴ 6561 should be multiplied by 3.

Product = 6561 × 3 = 19683

$$\begin{aligned}\Rightarrow \sqrt[3]{19683}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 3 \times 3 \times 3\\[1em] \Rightarrow 27\end{aligned}$$

Hence, 6561 should be multiplied by 3 to get the perfect cube 19683 and $\mathbf{\sqrt[3]{19683} = 27}$.

Finding prime factors of 8748 :

$$\begin{array}{l|r} 2 & 8748 \\ \hline 2 & 4374 \\ \hline 3 & 2187 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

8748 = 2 × 2 × (3 × 3 × 3) × (3 × 3 × 3) × 3

Since 2 × 2 × 3 is left after grouping the prime factors in triplets, so 8748 should be divided by 2 × 2 × 3 = 12.

Quotient = 8748 ÷ 12 = 729

$$\begin{aligned}\Rightarrow \sqrt[3]{729}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 3 \times 3\\[1em] \Rightarrow 9\end{aligned}$$

Hence, 8748 should be divided by 12 to get the perfect cube 729 and $\mathbf{\sqrt[3]{729} = 9}$.

Let the length of the side of the cubical box be $a$ cm.

We know that, volume of a cube = (side)³

Finding prime factors of 21952 :

$$\begin{array}{l|r} 2 & 21952 \\ \hline 2 & 10976 \\ \hline 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

21952 = (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7)

As, volume of the cube = 21952 cm³

$$\begin{aligned}\Rightarrow a^3 = 21952\\[1em] \Rightarrow a = \sqrt[3]{21952}\\[1em] \Rightarrow a = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow a = 2 \times 2 \times 7\\[1em] \Rightarrow a = 28\end{aligned}$$

Hence, the length of side of the box is 28 cm.

Let the three numbers be $3x, 4x$ and $5x$.

$$\begin{aligned}\Rightarrow 3x \times 4x \times 5x = 480\\[1em] \Rightarrow 60x^3 = 480\\[1em] \Rightarrow x^3 = \frac{480}{60}\\[1em] \Rightarrow x^3 = 8\\[1em] \Rightarrow x = \sqrt[3]{8}\\[1em] \Rightarrow x = \sqrt[3]{2 \times 2 \times 2}\\[1em] \Rightarrow x = 2\\[1em] \Rightarrow 3x = 3 \times 2 = 6\\[1em] \Rightarrow 4x = 4 \times 2 = 8\\[1em] \Rightarrow 5x = 5 \times 2 = 10\end{aligned}$$

Hence, the three numbers are 6, 8 and 10.

Let the two numbers be $4x$ and $5x$.

$$\begin{aligned}\Rightarrow (5x)^3 - (4x)^3 = 61\\[1em] \Rightarrow 125x^3 - 64x^3 = 61\\[1em] \Rightarrow 61x^3 = 61\\[1em] \Rightarrow x^3 = \frac{61}{61}\\[1em] \Rightarrow x^3 = 1\\[1em] \Rightarrow x = \sqrt[3]{1}\\[1em] \Rightarrow x = 1\\[1em] \Rightarrow 4x = 4 \times 1 = 4\\[1em] \Rightarrow 5x = 5 \times 1 = 5\end{aligned}$$

Hence, the two numbers are 4 and 5.

Cube root of the greater number = 8

∴ Greater number = 8³ = 512

Let the cube root of the smaller number be $x$.

∴ Smaller number = $x^3$

$$\begin{aligned}\Rightarrow 512 - x^3 = 387\\[1em] \Rightarrow x^3 = 512 - 387\\[1em] \Rightarrow x^3 = 125\\[1em] \Rightarrow x = \sqrt[3]{125}\\[1em] \Rightarrow x = \sqrt[3]{5 \times 5 \times 5}\\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, the cube root of the smaller number is 5.

(i) Cube of an even natural number is an even number.

(ii) Cubes of natural numbers are called perfect cubes.

(iii) Cube of a natural number whose unit digit is 7, ends with digit 3.

(iv) If a natural number has one zero at its end, then its cube will contain three zeros at the end.

(v) Finding cube root is the inverse operation of finding cube.

(vi) If a number can be expressed as the triplets of equal prime factors, then it is called perfect cube.

(i) False

Reason — Cubes of odd numbers are always odd. For example, 3³ = 27 and 5³ = 125, which are odd numbers.

(ii) True

Reason — If a number has one zero at its end, then its cube has three zeros at the end. So, the number of zeros at the end of a perfect cube is always a multiple of 3 and hence a perfect cube can never end with exactly two zeros.

(iii) False

Reason — If a number has 2 in the unit place, then its cube ends in 8. For example, 2³ = 8 and 12³ = 1728, which end with 8.

(iv) False

Reason — The smallest two digit number is 10 and 10³ = 1000, which is a four digit number. So, the cube of a two digit number always has four or more digits.

(v) True

Reason — 1³ = 1 and 2³ = 8, which are single digit numbers.

(vi) True

Reason — Let the number be $x$. On multiplying it by 2, we get $2x$.

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

So, the cube of the number gets multiplied by 8.

The cube of a negative integer is always negative.

Let the negative number be $(-a)$, where $a$ is a positive number.

Solving,

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

Since $a^3$ is positive, so -$a^3$ is negative.

Hence, Option 1 is the correct option.

If a number has 6 in the unit place, then its cube also ends in 6.

One's digit of 476 = 6

Cube of one's digit = 6 × 6 × 6 = 216

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (-8)^3\\[1em] \Rightarrow (-8) \times (-8) \times (-8)\\[1em] \Rightarrow -512\end{aligned}$$

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{-3}{7}\right)^3\\[1em] \Rightarrow \frac{(-3) \times (-3) \times (-3)}{7 \times 7 \times 7}\\[1em] \Rightarrow \frac{-27}{343}\end{aligned}$$

Hence, Option 2 is the correct option.

Finding prime factors of 1331 :

$$\begin{array}{l|r} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Expressing 1331 into prime factors, we have

1331 = 11 × 11 × 11

$$\begin{aligned}\Rightarrow \sqrt[3]{-1331}\\[1em] \Rightarrow -\sqrt[3]{1331}\\[1em] \Rightarrow -\sqrt[3]{11 \times 11 \times 11}\\[1em] \Rightarrow -11\end{aligned}$$

Hence, Option 3 is the correct option.

Finding prime factors of 2744 :

$$\begin{array}{l|r} 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 2744 into prime factors, we have

2744 = (2 × 2 × 2) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{2744}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 7\\[1em] \Rightarrow 14\end{aligned}$$

Hence, Option 4 is the correct option.

Converting mixed fraction to improper fraction:

$1\frac{127}{216} = \frac{343}{216}$

$$\begin{aligned}\Rightarrow \sqrt[3]{1\frac{127}{216}}\\[1em] \Rightarrow \sqrt[3]{\frac{343}{216}}\\[1em] \Rightarrow \frac{\sqrt[3]{343}}{\sqrt[3]{216}}\\[1em] \Rightarrow \frac{\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}\\[1em] \Rightarrow \frac{7}{2 \times 3}\\[1em] \Rightarrow \frac{7}{6}\\[1em] \Rightarrow 1\frac{1}{6}\end{aligned}$$

Hence, Option 3 is the correct option.

Finding prime factors of 192 :

$$\begin{array}{l|r} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

192 = (2 × 2 × 2) × (2 × 2 × 2) × 3

Since 3 is left after grouping the prime factors in triplets, so two more 3's are needed to complete the triplet.

∴ The required number = 3 × 3 = 9

Hence, Option 1 is the correct option.

Finding prime factors of 686 :

$$\begin{array}{l|r} 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

686 = 2 × (7 × 7 × 7)

Since 2 is left after grouping the prime factors in triplets, so 686 should be divided by 2.

686 ÷ 2 = 343 = (7 × 7 × 7), which is a perfect cube.

Hence, Option 2 is the correct option.

Let the length of the side of the cube be $a$ m.

We know that, volume of a cube = (side)³

Finding prime factors of 729 :

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Expressing 729 into prime factors, we have

729 = (3 × 3 × 3) × (3 × 3 × 3)

$$\begin{aligned}\Rightarrow a^3 = 729\\[1em] \Rightarrow a = \sqrt[3]{729}\\[1em] \Rightarrow a = \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow a = 3 \times 3\\[1em] \Rightarrow a = 9\end{aligned}$$

Hence, Option 3 is the correct option.

231 is an odd number and cubes of odd numbers are always odd. So, the cube of 231 is odd.

∴ Statement I is true.

If a number has 1 in its unit's place, then its cube also ends in 1, as 1 × 1 × 1 = 1.

∴ Statement II is true.

Hence, Option 3 is the correct option.

If $a$ is a positive integer, then $(-a)^3 = -a^3$, which is negative, while $a^2$ is positive.

So, $(-a)^3 \lt a^2$.

∴ Statement I is false.

If $a$ is a negative integer, then $(-a)$ is positive, so $(-a)^3$ is positive and $a^2$ is also positive.

Taking $a$ = -1, we get $(-a)^3$ = 1³ = 1 and $a^2$ = (-1)² = 1.

So, $(-a)^3 = a^2$, which means $(-a)^3 \gt a^2$ is not true for every negative integer.

∴ Statement II is false.

Hence, Option 4 is the correct option.

$(ab)^3 = (ab) × (ab) × (ab) = (a × a × a) × (b × b × b) = a^3b^3$

∴ Statement I is true.

We know that the cube root of the product of two numbers is equal to the product of their cube roots.

$$\begin{aligned}\Rightarrow \sqrt[3]{ab}\\[1em] \Rightarrow (ab)^{\frac{1}{3}}\\[1em] \Rightarrow a^{\frac{1}{3}} \times b^{\frac{1}{3}}\\[1em] \Rightarrow \sqrt[3]{a}\ \sqrt[3]{b}\end{aligned}$$

∴ Statement II is true.

Hence, Option 3 is the correct option.

Finding prime factors of 74088 :

$$\begin{array}{l|r} 2 & 74088 \\ \hline 2 & 37044 \\ \hline 2 & 18522 \\ \hline 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

74088 = (2 × 2 × 2) × (3 × 3 × 3) × (7 × 7 × 7)

Since 74088 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{74088}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 3 \times 7\\[1em] \Rightarrow 42\end{aligned}$$

Hence, 74088 is a perfect cube and it is the cube of 42.

Finding prime factors of 15625 :

$$\begin{array}{l|r} 5 & 15625 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

15625 = (5 × 5 × 5) × (5 × 5 × 5)

Since 15625 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{15625}\\[1em] \Rightarrow \sqrt[3]{(5 \times 5 \times 5) \times (5 \times 5 \times 5)}\\[1em] \Rightarrow 5 \times 5\\[1em] \Rightarrow 25\end{aligned}$$

Hence, 15625 is a perfect cube and it is the cube of 25.

Solving,

$$\begin{aligned}\Rightarrow (-17)^3\\[1em] \Rightarrow (-17) \times (-17) \times (-17)\\[1em] \Rightarrow -4913\end{aligned}$$

Hence, $\mathbf{(-17)^3 = -4913}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(-3\frac{4}{9}\Big)^3\\[1em] \Rightarrow \Big(\frac{-31}{9}\Big)^3\\[1em] \Rightarrow \frac{(-31) \times (-31) \times (-31)}{9 \times 9 \times 9}\\[1em] \Rightarrow \frac{-29791}{729}\\[1em] \Rightarrow -40\frac{631}{729}\end{aligned}$$

Hence, $\mathbf{\Big(-3\frac{4}{9}\Big)^3 = -40\frac{631}{729}}$.

Given number is 59319.

Finding prime factors of 59319 :

$$\begin{array}{l|r} 3 & 59319 \\ \hline 3 & 19773 \\ \hline 3 & 6591 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

Expressing 59319 into prime factors, we have

59319 = (3 × 3 × 3) × (13 × 13 × 13)

$$\begin{aligned}\Rightarrow \sqrt[3]{59319}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (13 \times 13 \times 13)}\\[1em] \Rightarrow 3 \times 13\\[1em] \Rightarrow 39\end{aligned}$$

Hence, cube root of 59319 = 39.

Given number is 21952.

Finding prime factors of 21952 :

$$\begin{array}{l|r} 2 & 21952 \\ \hline 2 & 10976 \\ \hline 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 21952 into prime factors, we have

21952 = (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{21952}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 7\\[1em] \Rightarrow 28\end{aligned}$$

Hence, cube root of 21952 = 28.

Finding prime factors of 9261 :

$$\begin{array}{l|r} 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Expressing 9261 into prime factors, we have

9261 = (3 × 3 × 3) × (7 × 7 × 7)

$$\begin{aligned}\Rightarrow \sqrt[3]{-9261}\\[1em] \Rightarrow -\sqrt[3]{9261}\\[1em] \Rightarrow -\sqrt[3]{(3 \times 3 \times 3) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow -(3 \times 7)\\[1em] \Rightarrow -21\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{-9261} = -21}$.

Converting mixed fraction to improper fraction:

$2\frac{43}{343} = \frac{729}{343}$

$$\begin{aligned}\Rightarrow \sqrt[3]{2\frac{43}{343}}\\[1em] \Rightarrow \sqrt[3]{\frac{729}{343}}\\[1em] \Rightarrow \frac{\sqrt[3]{729}}{\sqrt[3]{343}}\\[1em] \Rightarrow \frac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{7 \times 7 \times 7}}\\[1em] \Rightarrow \frac{3 \times 3}{7}\\[1em] \Rightarrow \frac{9}{7}\\[1em] \Rightarrow 1\frac{2}{7}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{2\frac{43}{343}} = 1\frac{2}{7}}$.

Converting decimal to fraction:

$0.216 = \frac{216}{1000}$

$$\begin{aligned}\Rightarrow \sqrt[3]{0.216}\\[1em] \Rightarrow \sqrt[3]{\frac{216}{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{216}}{\sqrt[3]{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow \frac{2 \times 3}{10}\\[1em] \Rightarrow \frac{6}{10}\\[1em] \Rightarrow 0.6\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{0.216} = 0.6}$.

Finding prime factors of 5184 :

$$\begin{array}{l|r} 2 & 5184 \\ \hline 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

5184 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 3

Since 3 is left after grouping the prime factors in triplets, so two more 3's are needed to complete the triplet.

∴ The required number = 3 × 3 = 9

Product = 5184 × 9 = 46656

$$\begin{aligned}\Rightarrow \sqrt[3]{46656}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow 36\end{aligned}$$

Hence, 5184 should be multiplied by 9 to get the perfect cube 46656 and $\mathbf{\sqrt[3]{46656} = 36}$.

Finding prime factors of 8788 :

$$\begin{array}{l|r} 2 & 8788 \\ \hline 2 & 4394 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

8788 = 2 × 2 × (13 × 13 × 13)

Since 2 × 2 is left after grouping the prime factors in triplets, so 8788 should be divided by 2 × 2 = 4.

Quotient = 8788 ÷ 4 = 2197

$$\begin{aligned}\Rightarrow \sqrt[3]{2197}\\[1em] \Rightarrow \sqrt[3]{13 \times 13 \times 13}\\[1em] \Rightarrow 13\end{aligned}$$

Hence, 8788 should be divided by 4 to get the perfect cube 2197 and $\mathbf{\sqrt[3]{2197} = 13}$.

Let the length of the side of the cube be $a$ cm.

We know that, volume of a cube = (side)³

Finding prime factors of 4096 :

$$\begin{array}{l|r} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array}$$

Expressing 4096 into prime factors, we have

4096 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

$$\begin{aligned}\Rightarrow a^3 = 4096\\[1em] \Rightarrow a = \sqrt[3]{4096}\\[1em] \Rightarrow a = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] \Rightarrow a = 2 \times 2 \times 2 \times 2\\[1em] \Rightarrow a = 16\end{aligned}$$

Hence, the side of the cube is 16 cm.

Total cost of painting = ₹1080

Rate of painting = ₹45 per sq. m

$$\begin{aligned}\Rightarrow \text{Outer surface area} = \frac{\text{Total cost}}{\text{Rate}}\\[1em] = \frac{1080}{45}\\[1em] = 24 \text{ sq. m}\end{aligned}$$

Let the length of the side of the cubical box be $a$ m.

Surface area of the cubical box = $6a^2$

$$\begin{aligned}\Rightarrow 6a^2 = 24\\[1em] \Rightarrow a^2 = \frac{24}{6}\\[1em] \Rightarrow a^2 = 4\\[1em] \Rightarrow a = \sqrt{4}\\[1em] \Rightarrow a = 2\end{aligned}$$

Volume of the cubical box = $a^3$ = 2 × 2 × 2 = 8

Hence, the volume of the cubical box is 8 m³.

Volume of the cube of side 4 cm = 4 × 4 × 4 = 64 cm³

Volume of the cube of side 6 cm = 6 × 6 × 6 = 216 cm³

Volume of 3 cuboids of dimensions 4 cm × 4 cm × 6 cm = 3 × (4 × 4 × 6) = 3 × 96 = 288 cm³

Volume of 3 cuboids of dimensions 4 cm × 6 cm × 6 cm = 3 × (4 × 6 × 6) = 3 × 144 = 432 cm³

Total volume = 64 + 216 + 288 + 432 = 1000 cm³

Let the length of the side of the new cube be $a$ cm.

$$\begin{aligned}\Rightarrow a^3 = 1000\\[1em] \Rightarrow a = \sqrt[3]{1000}\\[1em] \Rightarrow a = \sqrt[3]{10 \times 10 \times 10}\\[1em] \Rightarrow a = 10\end{aligned}$$

Since 1000 is a perfect cube, so the given cubes and cuboids can be arranged in the form of a big cube.

Hence, yes, a big cube can be formed and the length of its side is 10 cm.