NCERT Solutions for Class 8 Maths - Cubes and Cubes Roots

Exercise 7.1 

Question 1: Which of the following numbers are not perfect cubes: 

(i) 216 

(ii) 128 

(iii) 1000 

(iv) 100 

(v) 46656 

Answer : 

(i) 216

Prime factors of 216 = 2x2x2x3x3x3

Here all factors are in groups of 3’s [in triplets) Therefore, 216 is a perfect cube number.

(ii)

Prime factors of 128 =2x2x2x2x2x2x2

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

Therefore, 128 is not a perfect cube.

(iii)

Prime factors of 1000 =2x2x2x3x3x3

Here all factors appear in 3's group.

Therefore, 1000 is a perfect cube.

(iv)

Prime factors of 100 =2x2x5x5

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

Therefore, 100 is not a perfect cube.

(v)

Prime factors of 46656 = 2x2x2 x2x2x2x3x3x3x3x3x3

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 

Answer 2: 

(i)  

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)

Prime factors of 256 = 2 x 2 x 2 x 2x2x 2x2x2

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

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

(iii)

Prime factors of 72 = 2x2x2x3x3

Here 3 does not appear in 3's group.

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

(iv)

Prime factors of 675 =3x3x3x5x5

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

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

(v)

Prime factors of l00 = 2x2x5x5

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

Therefore 100 must be multiplied by 2 x 5 = 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 

Answer 3:

(i)

Prime factors of 81 = 3x3x3x3

Here one factor 3 is not grouped in triplets.

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

(ii)

Prime factors of 128 =2x2x2x2x2x2x2

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)

Prime factors of 135 = 3x3x3x5

Here one factor 5 does not appear in a triplet.

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

(iv)

Prime factors of 192 =2x2x2x2x2x2x3

Here one factor 3 does not appear in a triplet

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

(v)

Prime factors of 704 = 2x2x2x2x2x2x11

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? 

Answer 4: 

Given numbers = 5 x 2 x 5

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

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

Hence he needs 20 cuboids.

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

Answer 1: 

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.

Answer : 

(i) False

Since, 1³ '=1,3² = 27,5³ = 125,. ......... are all odd.

(ii) True

Since, a perfect cube ends with three zeroes.

e.g. 10³ = 1000,20³ =8000,30³ = 27000,...... soon

(iii) False

Since, 5² = 25,5³ = 125, 15² = 225,15³ = 3375 [Did not end with 2 5)

(iv) False

Since 12³ =1728 [Ends with 8]

And 22³ =10648 [Ends with 8]

(v) False

Since 10³ =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. 

Answer : 

We know that 10³ = 1000 and Possible cube of 11³ = 1331
Since, cube of unit's digit I³ = 1
Therefore, cube root of 1331 is 11,

4913

We know that 7³ = 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 13³ = 2197
And next number with 3 as ones digit 23³ =12167 Hence cube root of 12167 is 23.

32768

We know that 2³ = 8
Here in cube, ones digit is 8
Now next number with 2 as ones digit I2³ = 1728
And next number with 2 as ones digit 22³ = 10648
And next number with 2 as ones digit 32³ = 32768
Hence cube root of 32768 is 32.