Class 8 Maths Cubes and Cube Roots Worksheet - Free Printable

Problem Set Solution

Let's solve each problem step by step.

---

#### 1. Find the cube of 15.

The cube of a number \( n \) is given by \( n^3 \).

\[
15^3 = 15 \times 15 \times 15
\]

First, calculate \( 15 \times 15 \):

\[
15 \times 15 = 225
\]

Next, multiply the result by 15:

\[
225 \times 15 = 3375
\]

Thus, the cube of 15 is:

\[
\boxed{3375}
\]

---

#### 2. Is 3087 a perfect cube?

To determine if 3087 is a perfect cube, we need to find its prime factorization and check if all the exponents in the factorization are multiples of 3.

Step 1: Prime factorization of 3087

- 3087 is odd, so it is not divisible by 2.
- Sum of the digits of 3087 is \( 3 + 0 + 8 + 7 = 18 \), which is divisible by 3. So, 3087 is divisible by 3.

\[
3087 \div 3 = 1029
\]

- Now factorize 1029:
- Sum of the digits of 1029 is \( 1 + 0 + 2 + 9 = 12 \), which is divisible by 3. So, 1029 is divisible by 3.

\[
1029 \div 3 = 343
\]

- Now factorize 343:
- 343 is a known cube of 7 (\( 7^3 = 343 \)).

So, the prime factorization of 3087 is:

\[
3087 = 3^2 \times 7^3
\]

Step 2: Check if all exponents are multiples of 3

- The exponent of 3 is 2 (not a multiple of 3).
- The exponent of 7 is 3 (a multiple of 3).

Since not all exponents are multiples of 3, 3087 is not a perfect cube.

\[
\boxed{\text{No}}
\]

---

#### 3. If \( 72x \) is a perfect cube, then \( x = \) _______

To make \( 72x \) a perfect cube, we need to ensure that all the exponents in the prime factorization of \( 72x \) are multiples of 3.

Step 1: Prime factorization of 72

\[
72 = 2^3 \times 3^2
\]

Step 2: Determine what \( x \) should be

- The prime factorization of \( 72 \) is \( 2^3 \times 3^2 \).
- For \( 72x \) to be a perfect cube, the exponents of 2 and 3 in the factorization must be multiples of 3.
- Currently, the exponent of 2 is 3 (already a multiple of 3).
- The exponent of 3 is 2. To make it a multiple of 3, we need one more 3 (so the exponent becomes 3).

Thus, \( x \) must include at least one 3.

\[
x = 3
\]

Verify:

\[
72x = 72 \times 3 = 2^3 \times 3^2 \times 3 = 2^3 \times 3^3
\]

All exponents are now multiples of 3, so \( 72x \) is a perfect cube.

\[
\boxed{3}
\]

---

#### 4. Write the ones digit of the cube root of 2197.

To find the ones digit of the cube root of 2197, we observe the pattern in the ones digits of cubes.

| Number | Cube | Ones Digit of Cube |
|--------|------|---------------------|
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 2 | 8 | 8 |
| 3 | 27 | 7 |
| 4 | 64 | 4 |
| 5 | 125 | 5 |
| 6 | 216 | 6 |
| 7 | 343 | 3 |
| 8 | 512 | 2 |
| 9 | 729 | 9 |

The ones digit of 2197 is 7. From the table, the ones digit of the cube root of a number ending in 7 is 3.

Thus, the ones digit of the cube root of 2197 is:

\[
\boxed{3}
\]

---

#### 5. 17 is a cube root of _______.

If 17 is the cube root of a number, then the number is \( 17^3 \).

\[
17^3 = 17 \times 17 \times 17
\]

First, calculate \( 17 \times 17 \):

\[
17 \times 17 = 289
\]

Next, multiply the result by 17:

\[
289 \times 17 = 4913
\]

Thus, 17 is the cube root of:

\[
\boxed{4913}
\]

---

#### 6. The cube root of \( 2^3 \times 3^3 \times 3^3 \times 5^3 \) is _______.

Simplify the expression inside the cube root:

\[
2^3 \times 3^3 \times 3^3 \times 5^3 = 2^3 \times 3^{3+3} \times 5^3 = 2^3 \times 3^6 \times 5^3
\]

Take the cube root of each term:

\[
\sqrt[3]{2^3 \times 3^6 \times 5^3} = \sqrt[3]{2^3} \times \sqrt[3]{3^6} \times \sqrt[3]{5^3}
\]

\[
= 2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90
\]

Thus, the cube root is:

\[
\boxed{90}
\]

---

#### 7. Find the cube root of —

(i) 729

\[
729 = 9^3
\]

So, the cube root of 729 is:

\[
\sqrt[3]{729} = 9
\]

(ii) 64 × 27

First, calculate \( 64 \times 27 \):

\[
64 = 4^3 \quad \text{and} \quad 27 = 3^3
\]

\[
64 \times 27 = 4^3 \times 3^3 = (4 \times 3)^3 = 12^3
\]

So, the cube root of \( 64 \times 27 \) is:

\[
\sqrt[3]{64 \times 27} = 12
\]

(iii) 2744

\[
2744 = 14^3
\]

So, the cube root of 2744 is:

\[
\sqrt[3]{2744} = 14
\]

(iv) 64000

\[
64000 = 40^3
\]

So, the cube root of 64000 is:

\[
\sqrt[3]{64000} = 40
\]

(v) 1.331

\[
1.331 = 1.1^3
\]

So, the cube root of 1.331 is:

\[
\sqrt[3]{1.331} = 1.1
\]

---

#### 8. Find the cube root of \( 140 \times 2450 \).

First, simplify the product:

\[
140 \times 2450 = 140 \times (245 \times 10) = 1400 \times 245
\]

Factorize 1400 and 245:

\[
1400 = 2^3 \times 5^2 \times 7
\]
\[
245 = 5 \times 7^2
\]

Multiply the factorizations:

\[
1400 \times 245 = (2^3 \times 5^2 \times 7) \times (5 \times 7^2) = 2^3 \times 5^3 \times 7^3
\]

Take the cube root:

\[
\sqrt[3]{2^3 \times 5^3 \times 7^3} = 2 \times 5 \times 7 = 70
\]

Thus, the cube root is:

\[
\boxed{70}
\]

---

#### 9. Find the smallest number by which 256 must be multiplied so that the product is a perfect cube.

Step 1: Prime factorization of 256

\[
256 = 2^8
\]

Step 2: Make all exponents multiples of 3

- The exponent of 2 is 8. To make it a multiple of 3, we need to multiply by \( 2^1 \) (since \( 8 + 1 = 9 \)).

Thus, the smallest number to multiply by is:

\[
2
\]

Verify:

\[
256 \times 2 = 2^8 \times 2 = 2^9
\]

\( 2^9 \) is a perfect cube.

\[
\boxed{2}
\]

---

#### 10. If \( x^3 = 0.008 \), then \( x = \) _______

Given:

\[
x^3 = 0.008
\]

Take the cube root of both sides:

\[
x = \sqrt[3]{0.008}
\]

Notice that:

\[
0.008 = \frac{8}{1000} = \left( \frac{2}{10} \right)^3 = 0.2^3
\]

Thus:

\[
x = 0.2
\]

\[
\boxed{0.2}
\]

---

#### 11. Find the side of the cubical box whose volume is 9261 m³.

The volume of a cube is given by \( s^3 \), where \( s \) is the side length. Given:

\[
s^3 = 9261
\]

Take the cube root:

\[
s = \sqrt[3]{9261}
\]

Notice that:

\[
9261 = 21^3
\]

Thus:

\[
s = 21
\]

The side length of the cubical box is:

\[
\boxed{21}
\]

---

#### 12. If \( a^3 - 1 = 1330 \), then \( a = \) _______

Given:

\[
a^3 - 1 = 1330
\]

Add 1 to both sides:

\[
a^3 = 1331
\]

Take the cube root:

\[
a = \sqrt[3]{1331}
\]

Notice that:

\[
1331 = 11^3
\]

Thus:

\[
a = 11
\]

\[
\boxed{11}
\]

---

#### 13. Find the cube root of \( \frac{1331}{1728} \).

The cube root of a fraction is the cube root of the numerator divided by the cube root of the denominator:

\[
\sqrt[3]{\frac{1331}{1728}} = \frac{\sqrt[3]{1331}}{\sqrt[3]{1728}}
\]

Calculate each cube root:

\[
1331 = 11^3 \quad \Rightarrow \quad \sqrt[3]{1331} = 11
\]
\[
1728 = 12^3 \quad \Rightarrow \quad \sqrt[3]{1728} = 12
\]

Thus:

\[
\sqrt[3]{\frac{1331}{1728}} = \frac{11}{12}
\]

\[
\boxed{\frac{11}{12}}
\]

---

#### 14. If \( \sqrt[3]{\frac{x}{y}} = \frac{2}{3} \), then \( \frac{x}{y} = \) _______

Given:

\[
\sqrt[3]{\frac{x}{y}} = \frac{2}{3}
\]

Cube both sides:

\[
\left( \sqrt[3]{\frac{x}{y}} \right)^3 = \left( \frac{2}{3} \right)^3
\]

\[
\frac{x}{y} = \frac{8}{27}
\]

\[
\boxed{\frac{8}{27}}
\]

---

#### 15. Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 288. Find the numbers.

Let the three numbers be \( x \), \( 2x \), and \( 3x \). The sum of their cubes is given as 288:

\[
x^3 + (2x)^3 + (3x)^3 = 288
\]

Simplify each term:

\[
x^3 + 8x^3 + 27x^3 = 288
\]

Combine like terms:

\[
36x^3 = 288
\]

Solve for \( x^3 \):

\[
x^3 = \frac{288}{36} = 8
\]

Take the cube root:

\[
x = \sqrt[3]{8} = 2
\]

Thus, the three numbers are:

\[
x = 2, \quad 2x = 4, \quad 3x = 6
\]

The numbers are:

\[
\boxed{2, 4, 6}
\]

---

Final Answers:

1. \(\boxed{3375}\)
2. \(\boxed{\text{No}}\)
3. \(\boxed{3}\)
4. \(\boxed{3}\)
5. \(\boxed{4913}\)
6. \(\boxed{90}\)
7. (i) \(\boxed{9}\), (ii) \(\boxed{12}\), (iii) \(\boxed{14}\), (iv) \(\boxed{40}\), (v) \(\boxed{1.1}\)
8. \(\boxed{70}\)
9. \(\boxed{2}\)
10. \(\boxed{0.2}\)
11. \(\boxed{21}\)
12. \(\boxed{11}\)
13. \(\boxed{\frac{11}{12}}\)
14. \(\boxed{\frac{8}{27}}\)
15. \(\boxed{2, 4, 6}\)