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Access Answers to NCERT Class 8 Maths Chapter 7 – Cubes and Cube Roots Exercise 7.2 Page Number 275
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
Solution:
64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.
(ii) 512
Solution:
512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.
(iii) 10648
Solution:
10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.
Access Answers to NCERT Class 8 Maths Chapter 7 – Cubes and Cube Roots Exercise 7.2 Page Number 275
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
Solution:
64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.
(ii) 512
Solution:
512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.
(iii) 10648
Solution:
10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.
(iv) 27000
Solution:
27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 27000 = (2×2×2)×(3×3×3)×(5×5×5)
Here, 27000 can be grouped into triplets of equal factors,
∴ 27000 = (2×3×5) = 30
Hence, 30 is cube root of 27000.
(v) 15625
Solution:
15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors, 15625 = (5×5×5)×(5×5×5)
Here, 15625 can be grouped into triplets of equal factors,
∴ 15625 = (5×5) = 25
Hence, 25 is cube root of 15625.
(vi) 13824
Solution:
13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 13824 can be grouped into triplets of equal factors,
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is cube root of 13824.
(vii) 110592
Solution:
110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 110592 can be grouped into triplets of equal factors,
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is cube root of 110592.
(viii) 46656
Solution:
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 36 is cube root of 46656.
(ix) 175616
Solution:
175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)
Here, 175616 can be grouped into triplets of equal factors,
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is cube root of 175616.
(x) 91125
Solution:
91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3×3×5) = 45
Hence, 45 is cube root of 91125.
2. State true or false.
(i) Cube of any odd number is even.
Solution:
False
(ii) A perfect cube does not end with two zeros.
Solution:
True
(iii) If square of a number ends with 5, then its cube ends with 25.
Solution:
False
(iv) There is no perfect cube which ends with 8.
Solution:
False
(v) The cube of a two digit number may be a three digit number.
Solution:
False
(vi) The cube of a two digit number may have seven or more digits.
Solution:
False
(vii) The cube of a single digit number may be a single digit number.
Solution:
True
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
(i) By grouping the digits, we get 1 and 331
We know that, since, the unit digit of cube is 1, the unit digit of cube root is 1.
∴ We get 1 as unit digit of the cube root of 1331.
The cube of 1 matches with the number of second group.
∴ The ten’s digit of our cube root is taken as the unit place of smallest number.
We know that, the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ ∛1331 = 11
(ii) By grouping the digits, we get 4 and 913
We know that, since, the unit digit of cube is 3, the unit digit of cube root is 7.
∴ we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8.
Thus, 1 is taken as ten digit of cube root.
∴ ∛4913 = 17
(iii) By grouping the digits, we get 12 and 167.
We know that, since, the unit digit of cube is 7, the unit digit of cube root is 3.
∴ 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27 , 8 > 12 > 27.
Thus, 2 is taken as ten digit of cube root.
∴ ∛12167= 23
(iv) By grouping the digits, we get 32 and 768.
We know that, since, the unit digit of cube is 8, the unit digit of cube root is 2.
∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64.
Thus, 3 is taken as ten digit of cube root.
∴ ∛32768= 32