The document NCERT Solutions(Part- 2)- Cubes and Cube Roots Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**Exercise 7.1**

**Question 1. **Which of the following numbers are not perfect cubes?

(i) 216 (ii) 128 (iii) 1000

(iv) 100 (v) 46656

**Solution: **

**(i)** We have 216 = 2 * 2 * 2 * 3 * 3 * 3

Grouping the prime factors of 216 into triples, no factor is left over.

∴ 216 is a perfect cube.

**(ii)** We have 128 = 2 * 2 * 2 * 2 * 2 * 2 * 2

Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.

∴ 128 is not a perfect cube.

**(iii) **We have 1000 = 2 * 2 * 2 * 5 * 5 * 5

Grouping the prime factors of 1000 into triples, we are not left over with any factor.

∴ 1000 is a perfect cube.

**(iv)** We have 100 = 2 * 2 * 5 * 5

Grouping the prime factors into triples, we do not get any triples. Factors 2 * 2 and 5 * 5 are not in triples.

∴ 100 is not a perfect cube.

**(v) **We have 46656 = 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 3 * 3 * 3

Grouping the prime factors of 46656 in triples we are not left over with any prime factor.

∴ 46656 is a perfect cube.

**Question 2. **Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72

(iv) 675 (v) 100

**Solution: **

**(i)** We have 243 = 3 * 3 * 3 * 3 * 3

The prime factor 3 is not a group of three.

∴ 243 is not a perfect cube.

Now, [243] * 3 = [3 * 3 * 3 * 3 * 3] * 3

or 729 = 3 * 3 * 3 * 3 * 3 * 3

Now, 729 becomes a perfect cube.

Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

**(ii)** We have 256 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

Grouping the prime factors of 256 in triples, we are left over with 2 * 2.

∴ 256 is not a perfect cube.

Now, [256] * 2 = [2 * 2 * 2 * 2 * 2 * 2 * 2 * 2] * 2

or 512 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 i.e. 512 is a perfect cube.

Thus, the required smallest number is 2.

**(iii) **We have 72 = 2 * 2 * 2 * 3 * 3

Grouping the prime factors of 72 in triples, we are left over with 3 * 3.

∴ 72 is not a perfect cube.

Now, [72] * 3 = [2 * 2 * 2 * 3 * 3] * 3

or 216 = 2 * 2 * 2 * 3 * 3 * 3

i.e. 216 is a perfect cube.

∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

**(iv)** We have 675 = 3 * 3 * 3 * 5 * 5

Grouping the prime factors of 675 to triples, we are left over with 5 * 5.

∴ 675 is not a perfect cube.

Now, [675] * 5 = [3 * 3 * 3 * 5 * 5] * 5

or 3375 = 3 * 3 * 3 * 5 * 5 * 5

Now, 3375 is a perfect cube.

Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

**(v)** We have 100 = 2 * 2 * 5 * 5

The prime factor are not in the groups of triples.

∴ 100 is not a perfect cube.

Now [100] * 2 * 5 = [2 * 2 * 5 * 5] * 2 * 5

or [100] * 10 = 2 * 2 * 2 * 5 * 5 * 5

1000 = 2 * 2 * 2 * 5 * 5 * 5

Now, 1000 is a perfect cube.

Thus, the required smallest number is 10.

**Question 3.** Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135

(iv) 192 (v) 704

Solution:

**(i) **We have 81 = 3 * 3 * 3 * 3

Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is not a perfect cube.

Now, [81] /3 = [3 * 3 * 3 * 3] ÷ 3

or 27 = 3 * 3 * 3

i.e. 27 is a prefect cube

Thus, the required smallest number is 3.

**(ii)** We have 128 = 2 * 2 * 2 * 2 * 2 * 2 * 2

Grouping the prime factors of 128 into triples, we are left with 2.

∴ 128 is not a perfect cube

Now, [128] /2 = [2 * 2 * 2 * 2 * 2 * 2 * 2] ÷2

or 64 = 2 * 2 * 2 * 2 * 2 * 2

i.e. 64 is a perfect cube.

∴ The smallest required number is 2.

**(iii)** We have 135 = 3 * 3 * 3 * 5

Grouping the prime factors of 135 into triples, we are left over with 5.

∴ 135 is not a perfect cube

Now, [135] /5 = [3 * 3 * 3 * 5] ÷5

or 27 = 3 * 3 * 3

i.e. 27 is a perfect cube.

Thus, the required smallest number is 5.

**(iv)** We have 192 = 2 * 2 * 2 * 2 * 2 * 2 * 3

Grouping the prime factors of 192 into triples, 3 is left over.

∴ 192 is not a perfect cube.

Now, [192] ∏ 3 = [2 * 2 * 2 * 2 * 2 * 2 * 3] ÷3

or 64 = 2 * 2 * 2 * 2 * 2 * 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 3.

**(v)** We have 704 = 2 * 2 * 2 * 2 * 2 * 2 * 11

Grouping the prime factors of 704 into triples, 11 is left over.

∴ [704] /11 = [2 * 2 * 2 * 2 * 2 * 2 * 11] ÷11

or 64 = 2 * 2 * 2 * 2 * 2 * 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 11.

**Question 4.** Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

**Solution:** Sides of the cuboid are: 5 cm, 2 cm, 5 cm

∴ Volume of the cuboid = 5 cm * 2 cm * 5 cm

To form it as a cube its dimension should be in the group of triples.

∴ Volume of the required cube = [5 cm * 5 cm * 2 cm] * 5 cm * 2 cm * 2cm = [5 * 5 * 2 cm^{3}] = 20 cm^{3}

Thus, the required number of cuboids = 20

**Cube Roots**

Finding cube root is the inverse operation of finding cube.

Since, 4^{3 }= 64, so the cube root of 64 is 4.

The symbol for cube root is ∛

**E*amples: **

2^{3} = 8 → ∛8 = 2

3^{3} = 27 → ∛27 = 3

4^{3 }= 64 →∛64 = 4

5^{3} = 125 → ∛125 = 5

6^{3} = 216 → ∛ 216 = 6

7^{3} = 343 → ∛343 = 7

**Cube Root of a Cube Number through Estimation**

**Example 1.** Find the cube root of 614125 through estimation.

Solution: We use the following steps to find the cube root through estimation.

**I. **Form two groups three digits each starting from the right most”

**II.** 1st group (125) gives us the unit’s digit of cube root.

∵ Cube of a number ending in 5, also ends in 5.

∴ Unit’s digit of the cube root is 5.

**III. **2nd group (here 614) gives us the ten’s digit of the cube root.

Since, 8^{3} = 512 and 9^{3} = 729

Also 512 < 614 < 729

∴ We guess the ten’s digit of the cube root with the help of unit’s digit of 512.

We know that, if a number ends in 8, its cube will end in 2.

∴ Ten’s digit of the required cube root must be 8.

**Cube Root through Prime Factorisation Method**

**Example 1. **Find the cube root of 1728.

**Solution:** By prime factorisation, we have

**Example 2.** Find the cube root of 27000 by prime factorisation.

**Solution:**

∴

**Question: **State true or false: for any integer m, m^{2} < m^{3}. Why?

**Solution:** It is not always true.

For example, let m = –1

We have m^{2} = (–1)^{2} = 1

and m^{3} = (–1)^{3} = –1

∴ The above statement, i.e. m^{2} < m^{3} is false.

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