NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots

Exercise 7.1

Q.1) Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iv)100 (v) 46656
Sol.1) (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 × 2 × 5 × 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.

(v) 46656
Prime factors of 46656
= 2 × 2 × 2 × 2 × 2 × 2 × 3
× 3 × 3 × 3 × 3 × 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.

Q.2) obtain a perfect cube:
(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100
Sol.2) (i) 243
Prime factors of 243
= 3 × 3 × 3 × 3 × 3
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 𝑥 × 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
2 × 5 = 10 to make it a perfect cube.

Q.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
Sol.3) (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
= 2 × 2 × 2 × 2 × 2 × 2 × 2
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
= 3 × 3 × 3 × 5
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
= 2 × 2 × 2 × 2 × 2 × 2 × 3
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
= 2 × 2 × 2 × 2 × 2 × 2 × 11
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.

Q.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?
Sol.4) Given numbers = 5 × 2 × 5
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 ~2 × 5 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.

Exercise 7.2

Q.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
Sol.1) i) 64

Q.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.
Sol.2) (i) False
Since, 1³ = 1, 3³ = 2⁷, 5³ = 12⁵, … … are all odd.
(ii) True
Since, a perfect cube ends with three zeroes.
e.g. 10³ = 1000, 20³ = 8000, 30³ = 27000, … … so on
Since, 52 = 25, 53 = 125, 152 = 225, 153 = 3375 (Did not end with 25)
(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]

Q.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.
Sol.3) 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 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, one’s digit is 7
Now next number with 3 as one’s digit 133 = 2197
And next number with 3 as one’s digit 233 = 12167
Hence cube root of 12167 is 23.
32768
We know that 3 2 = 8
Here in cube, one’s digit is 8
Now next number with 2 as ones digit 123 = 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.