Chapter 6 Cubes and Cube Roots

Solution:

A number is a perfect cube if its prime factors can be grouped in triplets.

Let's find the prime factors of 243.

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

The prime factorization of 243 is $3 \times 3 \times 3 \times 3 \times 3$.

We can write this as $243 = 3^5$.

To be a perfect cube, the exponent of each prime factor in the factorization must be a multiple of 3.

In this case, the only prime factor is 3, and its exponent is 5.

Since 5 is not a multiple of 3, the prime factors cannot be grouped into complete triplets.

$243 = \underbrace{3 \times 3 \times 3}_{\text{Triplet}} \times 3 \times 3 = 3^3 \times 3^2$.

There are two factors of 3 remaining after forming one triplet.

Therefore, 243 is not a perfect cube.

Solution:

A number is a perfect cube if its prime factors can be grouped in triplets.

Let's find the prime factors of 392.

$$\begin{array}{c|cc} 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

The prime factorization of 392 is $2 \times 2 \times 2 \times 7 \times 7$.

We can write this as $392 = 2^3 \times 7^2$.

To be a perfect cube, the exponent of each prime factor must be a multiple of 3.

In the prime factorization of 392, the exponent of 2 is 3 (which is a multiple of 3), but the exponent of 7 is 2 (which is not a multiple of 3).

$392 = \underbrace{2 \times 2 \times 2}_{\text{Triplet}} \times 7 \times 7 = 2^3 \times 7^2$.

Since the prime factor 7 does not appear in a triplet, 392 is not a perfect cube.

---

To make 392 a perfect cube, we need to multiply it by the smallest natural number that will complete the triplet of the prime factor 7.

We have $7^2$. To make it $7^3$ (which has an exponent that is a multiple of 3), we need to multiply by $7^1$, which is 7.

If we multiply 392 by 7, the product will be:

$392 \times 7 = (2^3 \times 7^2) \times 7 = 2^3 \times 7^{2+1} = 2^3 \times 7^3$

This can be written as $(2 \times 7)^3 = 14^3$.

Since the exponents of all prime factors (2 and 7) are now multiples of 3, the product $2^3 \times 7^3$ is a perfect cube.

The smallest natural number by which 392 must be multiplied to get a perfect cube is 7.

Solution:

A number is a perfect cube if its prime factors can be grouped in triplets.

Let's find the prime factors of 53240.

$$\begin{array}{c|cc} 2 & 53240 \\ \hline 2 & 26620 \\ \hline 2 & 13310 \\ \hline 5 & 6655 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

The prime factorization of 53240 is $2 \times 2 \times 2 \times 5 \times 11 \times 11 \times 11$.

We can write this as $53240 = 2^3 \times 5^1 \times 11^3$.

To be a perfect cube, the exponent of each prime factor must be a multiple of 3.

$53240 = \underbrace{2 \times 2 \times 2}_{\text{Triplet}} \times 5 \times \underbrace{11 \times 11 \times 11}_{\text{Triplet}} = 2^3 \times 5^1 \times 11^3$.

In the prime factorization of 53240, the exponents of 2 and 11 are 3 (which is a multiple of 3), but the exponent of 5 is 1 (which is not a multiple of 3).

Since the prime factor 5 does not appear in a triplet, 53240 is not a perfect cube.

---

To make the quotient a perfect cube after division, we need to eliminate the prime factors that are not part of a triplet.

The prime factor 5 has an exponent of 1, which is not a multiple of 3.

To make the exponent of 5 a multiple of 3 (in this case, $5^0$ by division), we need to divide by $5^1$, which is 5.

If we divide 53240 by 5, the quotient will be:

$\frac{53240}{5} = \frac{2^3 \times 5^1 \times 11^3}{5^1} = 2^3 \times 5^{1-1} \times 11^3 = 2^3 \times 5^0 \times 11^3 = 2^3 \times 1 \times 11^3 = 2^3 \times 11^3$

This can be written as $(2 \times 11)^3 = 22^3$.

Since the exponents of all prime factors (2 and 11) in the quotient are now multiples of 3, the quotient $2^3 \times 11^3$ is a perfect cube.

The smallest natural number by which 53240 must be divided to get a perfect cube is 5.

Solution:

A number is a perfect cube if its prime factors can be grouped in triplets.

Let's find the prime factors of 1188.

$$\begin{array}{c|cc} 2 & 1188 \\ \hline 2 & 594 \\ \hline 3 & 297 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

The prime factorization of 1188 is $2 \times 2 \times 3 \times 3 \times 3 \times 11$.

We can write this as $1188 = 2^2 \times 3^3 \times 11^1$.

To be a perfect cube, the exponent of each prime factor must be a multiple of 3.

$1188 = 2^2 \times \underbrace{3 \times 3 \times 3}_{\text{Triplet}} \times 11^1 = 2^2 \times 3^3 \times 11^1$.

In the prime factorization of 1188, the exponent of 3 is 3 (which is a multiple of 3), but the exponents of 2 (which is 2) and 11 (which is 1) are not multiples of 3.

Since the prime factors 2 and 11 do not appear in complete triplets, 1188 is not a perfect cube.

---

To make the quotient a perfect cube after division, we need to eliminate the prime factors that are not part of a triplet.

The prime factor 2 has an exponent of 2. To make this a multiple of 3 (ideally $2^0$ by division), we need to divide by $2^2$.

The prime factor 11 has an exponent of 1. To make this a multiple of 3 (ideally $11^0$ by division), we need to divide by $11^1$.

The smallest number to divide 1188 by is the product of these "extra" prime factors raised to their current powers: $2^2 \times 11^1$.

Smallest number to divide by $= 2^2 \times 11^1 = 4 \times 11 = 44$.

If we divide 1188 by 44, the quotient will be:

$\frac{1188}{44} = \frac{2^2 \times 3^3 \times 11^1}{2^2 \times 11^1} = 3^3$.

The quotient is $3^3 = 27$, which is a perfect cube ($27 = 3 \times 3 \times 3$).

The smallest natural number by which 1188 must be divided to get a perfect cube is 44.

Solution:

A number is a perfect cube if its prime factors can be grouped in triplets.

Let's find the prime factors of 68600.

$$\begin{array}{c|cc} 2 & 68600 \\ \hline 2 & 34300 \\ \hline 2 & 17150 \\ \hline 5 & 8575 \\ \hline 5 & 1715 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

The prime factorization of 68600 is $2 \times 2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 7$.

We can write this as $68600 = 2^3 \times 5^2 \times 7^3$.

To be a perfect cube, the exponent of each prime factor must be a multiple of 3.

In the prime factorization of 68600, the exponents of 2 and 7 are 3 (which are multiples of 3), but the exponent of 5 is 2 (which is not a multiple of 3).

Since the prime factor 5 does not appear in a complete triplet ($5 \times 5$ is missing one factor of 5 to form a triplet), 68600 is not a perfect cube.

---

To make 68600 a perfect cube, we need to multiply it by the smallest natural number that will complete the triplet of the prime factor 5.

We have $5^2$. To make it $5^3$ (which has an exponent that is a multiple of 3), we need to multiply by $5^{3-2} = 5^1$, which is 5.

If we multiply 68600 by 5, the product will be:

$68600 \times 5 = (2^3 \times 5^2 \times 7^3) \times 5 = 2^3 \times 5^{2+1} \times 7^3 = 2^3 \times 5^3 \times 7^3$

This can be written as $(2 \times 5 \times 7)^3 = 70^3$.

Since the exponents of all prime factors (2, 5, and 7) are now multiples of 3, the product $2^3 \times 5^3 \times 7^3$ is a perfect cube.

The smallest natural number by which 68600 must be multiplied to get a perfect cube is 5.

Solution:

A number is a perfect cube if, in its prime factorization, all the exponents of the prime factors are multiples of 3.

---

(i) 216

Let's find the prime factors of 216.

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$.

The exponents of the prime factors 2 and 3 are 3, which is a multiple of 3.

Therefore, 216 is a perfect cube ($216 = 6^3$).

---

(ii) 128

Let's find the prime factors of 128.

$\begin{array}{c|cc} 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

The prime factorization of 128 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.

The exponent of the prime factor 2 is 7, which is not a multiple of 3.

Therefore, 128 is not a perfect cube.

---

(iii) 1000

Let's find the prime factors of 1000.

$\begin{array}{c|cc} 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 1000 is $2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3$.

The exponents of the prime factors 2 and 5 are 3, which is a multiple of 3.

Therefore, 1000 is a perfect cube ($1000 = 10^3$).

---

(iv) 100

Let's find the prime factors of 100.

$\begin{array}{c|cc} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 100 is $2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$.

The exponents of the prime factors 2 and 5 are 2, which is not a multiple of 3.

Therefore, 100 is not a perfect cube.

---

(v) 46656

Let's find the prime factors of 46656.

$\begin{array}{c|cc} 2 & 46656 \\ \hline 2 & 23328 \\ \hline 2 & 11664 \\ \hline 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 46656 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^6 \times 3^6$.

The exponents of the prime factors 2 and 3 are 6, which is a multiple of 3.

Therefore, 46656 is a perfect cube ($46656 = 36^3$).

---

The numbers that are not perfect cubes are 128 and 100.

Solution:

To find the smallest number by which a given number must be multiplied to obtain a perfect cube, we find the prime factorization of the number. For the product to be a perfect cube, the exponent of each prime factor in the factorization must be a multiple of 3. We identify the prime factors whose exponents are not multiples of 3 and multiply by the smallest power of these primes needed to make their exponents multiples of 3.

---

(i) 243

Prime factorization of 243:

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

$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$.

The exponent of 3 is 5, which is not a multiple of 3.

To make the exponent a multiple of 3 (the next multiple is 6), we need to multiply by $3^{6-5} = 3^1 = 3$.

The smallest number to multiply by is 3.

---

(ii) 256

Prime factorization of 256:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8$.

The exponent of 2 is 8, which is not a multiple of 3.

To make the exponent a multiple of 3 (the next multiple is 9), we need to multiply by $2^{9-8} = 2^1 = 2$.

The smallest number to multiply by is 2.

---

(iii) 72

Prime factorization of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

The exponent of 2 is 3, which is a multiple of 3.

The exponent of 3 is 2, which is not a multiple of 3.

To make the exponent of 3 a multiple of 3 (the next multiple is 3), we need to multiply by $3^{3-2} = 3^1 = 3$.

The smallest number to multiply by is 3.

---

(iv) 675

Prime factorization of 675:

$\begin{array}{c|cc} 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5^2$.

The exponent of 3 is 3, which is a multiple of 3.

The exponent of 5 is 2, which is not a multiple of 3.

To make the exponent of 5 a multiple of 3 (the next multiple is 3), we need to multiply by $5^{3-2} = 5^1 = 5$.

The smallest number to multiply by is 5.

---

(v) 100

Prime factorization of 100:

$\begin{array}{c|cc} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$.

The exponent of 2 is 2, which is not a multiple of 3.

The exponent of 5 is 2, which is not a multiple of 3.

To make the exponent of 2 a multiple of 3, we need to multiply by $2^{3-2} = 2^1 = 2$.

To make the exponent of 5 a multiple of 3, we need to multiply by $5^{3-2} = 5^1 = 5$.

The smallest number to multiply by is $2 \times 5 = 10$.

Solution:

To find the smallest number by which a given number must be divided to obtain a perfect cube, we find the prime factorization of the number. For the quotient to be a perfect cube, the exponent of each prime factor in the factorization must be a multiple of 3. We identify the prime factors whose exponents are not multiples of 3 and divide by these factors raised to their current powers to eliminate them.

---

(i) 81

Prime factorization of 81:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$81 = 3 \times 3 \times 3 \times 3 = 3^4$.

The exponent of 3 is 4, which is not a multiple of 3.

We have one triplet of 3 ($3^3$) and one extra factor of 3 ($3^1$). To obtain a perfect cube by division, we must remove the extra factors.

$81 = \underbrace{3 \times 3 \times 3}_{\text{Triplet}} \times 3$.

The factor that is not part of a triplet is $3^1 = 3$.

The smallest number to divide by is 3.

---

(ii) 128

Prime factorization of 128:

$\begin{array}{c|cc} 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.

The exponent of 2 is 7, which is not a multiple of 3.

We have two triplets of 2 ($2^3 \times 2^3$) and one extra factor of 2 ($2^1$). To obtain a perfect cube by division, we must remove the extra factors.

$128 = \underbrace{2 \times 2 \times 2}_{\text{Triplet}} \times \underbrace{2 \times 2 \times 2}_{\text{Triplet}} \times 2$.

The factor that is not part of a triplet is $2^1 = 2$.

The smallest number to divide by is 2.

---

(iii) 135

Prime factorization of 135:

$\begin{array}{c|cc} 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$135 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5^1$.

The exponent of 3 is 3, which is a multiple of 3.

The exponent of 5 is 1, which is not a multiple of 3.

The factor that is not part of a triplet is $5^1 = 5$.

The smallest number to divide by is 5.

---

(iv) 192

Prime factorization of 192:

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

$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$.

The exponent of 2 is 6, which is a multiple of 3.

The exponent of 3 is 1, which is not a multiple of 3.

The factor that is not part of a triplet is $3^1 = 3$.

The smallest number to divide by is 3.

---

(v) 704

Prime factorization of 704:

$\begin{array}{c|cc} 2 & 704 \\ \hline 2 & 352 \\ \hline 2 & 176 \\ \hline 2 & 88 \\ \hline 2 & 44 \\ \hline 2 & 22 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^6 \times 11^1$.

The exponent of 2 is 6, which is a multiple of 3.

The exponent of 11 is 1, which is not a multiple of 3.

The factor that is not part of a triplet is $11^1 = 11$.

The smallest number to divide by is 11.

Solution:

The dimensions of the cuboid are 5 cm, 2 cm, and 5 cm.

The volume of one cuboid is the product of its dimensions: $5 \times 2 \times 5$ cubic cm.

Volume of one cuboid $= 5 \times 2 \times 5 = 50$ cubic cm.

---

To form a cube from these cuboids, the total volume of the combined shape must be a perfect cube.

Let the number of cuboids needed be $N$.

The total volume of $N$ cuboids will be $N \times (5 \times 2 \times 5)$.

This total volume must be equal to the volume of a cube, say with side length $S$. So, $S^3 = N \times (5 \times 2 \times 5)$.

Let's look at the prime factorization of the volume of one cuboid:

$5 \times 2 \times 5 = 2^1 \times 5^2$.

So, $S^3 = N \times 2^1 \times 5^2$.

For $S^3$ to be a perfect cube, the exponent of each prime factor in its prime factorization must be a multiple of 3.

Currently, the prime factors in $2^1 \times 5^2$ are 2 (with exponent 1) and 5 (with exponent 2).

To make the exponent of 2 a multiple of 3, we need $2^{3-1} = 2^2$ more factors of 2.

To make the exponent of 5 a multiple of 3, we need $5^{3-2} = 5^1$ more factors of 5.

The smallest number $N$ that provides these missing factors is $2^2 \times 5^1$.

$N = 2^2 \times 5^1 = 4 \times 5 = 20$.

---

If we use 20 cuboids, the total volume will be:

$20 \times (5 \times 2 \times 5) = (2^2 \times 5^1) \times (2^1 \times 5^2) = 2^{2+1} \times 5^{1+2} = 2^3 \times 5^3 = (2 \times 5)^3 = 10^3$.

The total volume is $10^3 = 1000$ cubic cm, which is the volume of a cube with side length 10 cm.

This can be achieved by arranging the cuboids: we need 2 cuboids along one 5cm side ($2 \times 5 = 10$), 5 cuboids along the 2cm side ($5 \times 2 = 10$), and 2 cuboids along the other 5cm side ($2 \times 5 = 10$). The total number of cuboids used is $2 \times 5 \times 2 = 20$.

---

Therefore, Parikshit will need 20 such cuboids to form a cube.

Solution:

To find the cube root of 8000 by prime factorization, we list the prime factors of 8000 and group them into triplets.

Let's find the prime factors of 8000.

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

The prime factorization of 8000 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$.

We can group the factors in triplets:

$8000 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5)$

$8000 = 2^3 \times 2^3 \times 5^3$

Using the property $(a \times b)^m = a^m \times b^m$, we can write:

$8000 = (2 \times 2 \times 5)^3 = (4 \times 5)^3 = 20^3$.

The cube root of 8000 is found by taking one factor from each triplet:

$\sqrt[3]{8000} = \sqrt[3]{2^3 \times 2^3 \times 5^3} = 2 \times 2 \times 5 = 20$.

Alternatively, $\sqrt[3]{2^3 \times 2^3 \times 5^3} = \sqrt[3]{(2 \times 2 \times 5)^3} = 2 \times 2 \times 5 = 20$.

---

Thus, the cube root of 8000 is 20.

$\sqrt[3]{8000} = 20$.

Solution:

To find the cube root of 13824 using the prime factorization method, we follow these steps:

1. Find the prime factors of the number.

2. Group the prime factors into triplets.

3. Take one factor from each triplet and multiply them to find the cube root.

---

Let's find the prime factors of 13824.

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

The prime factorization of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

We can write this in exponential form as $13824 = 2^9 \times 3^3$.

---

Now, we group the prime factors into triplets:

$13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

This can be written as $13824 = 2^3 \times 2^3 \times 2^3 \times 3^3$.

Using the property $(a \times b \times c \times d)^3 = a^3 \times b^3 \times c^3 \times d^3$, we have:

$13824 = (2 \times 2 \times 2 \times 3)^3 = (8 \times 3)^3 = 24^3$.

---

To find the cube root, we take one factor from each triplet of prime factors:

$\sqrt[3]{13824} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}$

$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 8 \times 3 = 24$.

Alternatively, using the exponential form: $\sqrt[3]{13824} = \sqrt[3]{2^9 \times 3^3} = 2^{9/3} \times 3^{3/3} = 2^3 \times 3^1 = 8 \times 3 = 24$.

---

Thus, the cube root of 13824 is 24.

$\sqrt[3]{13824} = 24$.

Solution:

To find the cube root of a number using the prime factorization method, we follow these steps:

1. Find the prime factorization of the given number.

2. Group the identical prime factors in triplets.

3. For each triplet, take one factor.

4. Multiply the factors taken from each triplet. This product is the cube root of the number.

---

(i) 64

Prime factorization of 64:

The prime factorization of 64 is $2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2)$.

Taking one factor from each triplet: $2 \times 2 = 4$.

Therefore, $\sqrt[3]{64} = 4$.

---

(ii) 512

Prime factorization of 512:

The prime factorization of 512 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)$.

Taking one factor from each triplet: $2 \times 2 \times 2 = 8$.

Therefore, $\sqrt[3]{512} = 8$.

---

(iii) 10648

Prime factorization of 10648:

The prime factorization of 10648 is $2 \times 2 \times 2 \times 11 \times 11 \times 11$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (11 \times 11 \times 11)$.

Taking one factor from each triplet: $2 \times 11 = 22$.

Therefore, $\sqrt[3]{10648} = 22$.

---

(iv) 27000

Prime factorization of 27000:

The prime factorization of 27000 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5)$.

Taking one factor from each triplet: $2 \times 3 \times 5 = 30$.

Therefore, $\sqrt[3]{27000} = 30$.

---

(v) 15625

Prime factorization of 15625:

The prime factorization of 15625 is $5 \times 5 \times 5 \times 5 \times 5 \times 5$.

Grouping the factors in triplets: $(5 \times 5 \times 5) \times (5 \times 5 \times 5)$.

Taking one factor from each triplet: $5 \times 5 = 25$.

Therefore, $\sqrt[3]{15625} = 25$.

---

(vi) 13824

Prime factorization of 13824:

The prime factorization of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$.

Taking one factor from each triplet: $2 \times 2 \times 2 \times 3 = 8 \times 3 = 24$.

Therefore, $\sqrt[3]{13824} = 24$.

---

(vii) 110592

Prime factorization of 110592:

The prime factorization of 110592 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$. There are 12 factors of 2 and 3 factors of 3.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$.

Taking one factor from each triplet: $2 \times 2 \times 2 \times 2 \times 3 = 16 \times 3 = 48$.

Therefore, $\sqrt[3]{110592} = 48$.

---

(viii) 46656

Prime factorization of 46656:

The prime factorization of 46656 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)$.

Taking one factor from each triplet: $2 \times 2 \times 3 \times 3 = 4 \times 9 = 36$.

Therefore, $\sqrt[3]{46656} = 36$.

---

(ix) 175616

Prime factorization of 175616:

The prime factorization of 175616 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7$.

Grouping the factors in triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)$.

Taking one factor from each triplet: $2 \times 2 \times 2 \times 7 = 8 \times 7 = 56$.

Therefore, $\sqrt[3]{175616} = 56$.

---

(x) 91125

Prime factorization of 91125:

The prime factorization of 91125 is $3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$.

Grouping the factors in triplets: $(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5)$.

Taking one factor from each triplet: $3 \times 3 \times 5 = 9 \times 5 = 45$.

Therefore, $\sqrt[3]{91125} = 45$.

Solution:

---

(i) Cube of any odd number is even.

The cube of an odd number is always odd. For example, $3^3 = 27$ (odd), $5^3 = 125$ (odd).

False

---

(ii) A perfect cube does not end with two zeros.

For a number to be a perfect cube and end with zeros, it must end with a number of zeros that is a multiple of 3 (e.g., 1000, 1000000). A number ending with two zeros means it is a multiple of 100 ($10^2$), which has prime factors $2^2 \times 5^2$. For it to be a perfect cube, the exponents must be multiples of 3. Since the exponents of 2 and 5 are 2 (not a multiple of 3), such a number cannot be a perfect cube.

True

---

(iii) If square of a number ends with 5, then its cube ends with 25.

If the square of a number ends with 5, the number itself must end with 5. Let's check cubes of numbers ending in 5: $5^3 = 125$ (ends in 25), $15^3 = 3375$ (ends in 75), $25^3 = 15625$ (ends in 25). The statement is not always true.

False

---

(iv) There is no perfect cube which ends with 8.

Let's check the unit digits of cubes: $2^3=8$, $12^3=1728$. Perfect cubes can end with the digit 8.

False

---

(v) The cube of a two digit number may be a three digit number.

The smallest two-digit number is 10. $10^3 = 1000$, which is a four-digit number. Any two-digit number is greater than or equal to 10, so its cube will be greater than or equal to 1000, which has at least four digits.

False

---

(vi) The cube of a two digit number may have seven or more digits.

The largest two-digit number is 99. $99^3 = 970299$, which has six digits. The cube of any two-digit number will be between $10^3 = 1000$ (4 digits) and $99^3 = 970299$ (6 digits). Thus, the cube of a two-digit number can have 4, 5, or 6 digits, but not 7 or more.

False

---

(vii) The cube of a single digit number may be a single digit number.

Let's check the cubes of single-digit numbers: $0^3 = 0$ (single digit), $1^3 = 1$ (single digit), $2^3 = 8$ (single digit). Since there are single-digit numbers whose cubes are also single-digit numbers, the statement is true.

True