### 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)216

Prime factors of 216 = 2×2×2×3×3×3

Here all factors are in groups of 3’s (in triplets)

Therefore, 216 is a perfect cube number.

(ii) 128

Prime factors of 128

= 2×2×2×2×2×2×2

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 is not a perfect cube.

(iii) 1000

Prime factors of 1000 = 2×2×2×3×3×3

Here all factors appear in 3’s group.

Therefore, 1000 is a perfect cube.

(iv) 100

Prime factors of 100 = 2 x 2 x 5 x 5

Here all factors do not appear in 3’s group.

Therefore, 100 is not a perfect cube.

(v) 46656

Prime factors of 46656

= **2x2x2x2x2x2x3x3x3x3x3x3**

Here all factors appear in 3’s group.

Therefore, 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)243

Prime factors of 243 = **3x3x3x3x3**

Here 3 does not appear in 3’s group.

Therefore, 243 must be multiplied by 3 to make it a perfect cube.

(ii) 256

Prime factors of 256

= 2×2×2×2×2×2×2×2

Here one factor 2 is required to make a 3’s group.

Therefore, 256 must be multiplied by 2 to make it a perfect cube.

(iii) 72

Prime factors of 72 = 2×2×2×3×3

Here 3 does not appear in 3’s group.

Therefore, 72 must be multiplied by 3 to make it a perfect cube.

(iv) 675

Prime factors of 675 = 3 × 3 × 3 × 5 × 5

Here factor 5 does not appear in 3’s group.

Therefore 675 must be multiplied by 3 to make it a perfect cube.

(v) 100

Prime factors of 100 =2 × 2 × 5 × 5

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by = 10 to make it a perfect cube.

**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)81

Prime factors of 81 = 3 × 3 × 3 × 3

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

(ii) 128

Prime factors of 128 =

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 must be divided by 2 to make it a perfect cube.

(iii) 135

Prime factors of 135 =

Here one factor 5 does not appear in a triplet.

Therefore, 135 must be divided by 5 to make it a perfect cube.

(iv) 192

Prime factors of 192 =

Here one factor 3 does not appear in a triplet.

Therefore, 192 must be divided by 3 to make it a perfect cube.

(v) 704

Prime factors of 704

=

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

**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?

Ans. Given numbers =

Since, Factors of 5 and 2 both are not in group of three.

Therefore, the number must be multiplied by 2 × 5 × 2 = 20 to make it a perfect cube.

Hence he needs 20 cuboids.

### NCERT Solutions for Class 8 Maths Exercise 7.2

**Question 1.**

Find the cube root of each of the following numbers by prime factorization method:

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

**Solution :** (i) 64

= 4

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

=5 × 5

= 25

(vi) 13824

=

= 24

(vii) 110592

= **2×2×2×2×3**

= 48

(viii) 46656

= 36

(ix) 175616

= 56

(x) 91125

**Question 2.**

State true or false:

(i) Cube of any odd number is even.

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

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

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

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

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

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

**Solution :** (i) False

Since, are all odd.

(ii) True

Since, a perfect cube ends with three zeroes. e.g.so on

(iii) False

Since,

(Did not end with 25)

(iv) False

Since = 1728

[Ends with 8]

And = 10648

[Ends with 8]

(v) False Since = 1000

[Four digit number]

And **11³ =** 1331

[Four digit number]

(vi) False Since **99³ **= 970299

[Six digit number]

(vii) True

**1³** = 1

[Single digit number]** 2³** = 8

[Single digit number]

**Question 3.**

You are told that 1,331 is a perfect cube. Can you guess with factorization what is its cube root? Similarly guess the cube roots of 4913, 12167, 32768.**Solution :**

We know that **10³ **= 1000 and Possible cube of **11³ **= 1331

Since, cube of unit’s digit **1³** = 1

Therefore, cube root of 1331 is 11.

4913

We know that 7**³ **= 343

Next number comes with 7 as unit place **17³ **= 4913

Hence, cube root of 4913 is 17.

12167

We know that **3³ **= 27

Here in cube, ones digit is 7

Now next number with 3 as ones digit

13**³ **= 2197

And next number with 3 as ones digit

23**³** = 12167

Hence cube root of 12167 is 23.

32768

We know that 2**³ **= 8

Here in cube, ones digit is 8

Now next number with 2 as ones digit

12**³ **= 1728

And next number with 2 as ones digit

22**³ **= 10648

And next number with 2 as ones digit

32**³** = 32768

Hence cube root of 32768 is 32.