NCERT Solutions for Class 8 Maths – Ganita Prakash
Exercise 1.2
1. Find the cube roots of 27000 and 10648.
(i) Cube root of 27000
Prime factorisation:
27000 = 27 × 1000
= 33 × 103
= 33 × 23 × 53
∛27000 = ∛(33 × 23 × 53)
= 3 × 2 × 5
= 30
Therefore, ∛27000 = 30.
(ii) Cube root of 10648
Prime factorisation:
10648 = 2 × 2 × 2 × 11 × 11 × 11
= 23 × 113
∛10648 = ∛(23 × 113)
= 2 × 11
= 22
Therefore, ∛10648 = 22.
2. What number will you multiply by 1323 to make it a cube number?
Prime factorisation of 1323:
1323 = 3 × 441
= 3 × 3 × 147
= 3 × 3 × 3 × 49
= 33 × 72
For a perfect cube, the powers of all prime factors must be multiples of 3.
- Power of 3 is 3 → already a multiple of 3
- Power of 7 is 2 → needs one more 7
So, we multiply 1323 by 7.
1323 × 7 = 9261 = 213
Therefore, the required number is 7.
3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
An odd number multiplied by an odd number remains odd.
So, odd × odd × odd = odd.
Therefore, the statement is False.
(ii) There is no perfect cube that ends with 8.
Example: 23 = 8 and 123 = 1728.
Both perfect cubes end with 8.
Therefore, the statement is False.
(iii) The cube of a 2-digit number may be a 3-digit number.
Take 10 as a 2-digit number.
103 = 1000, which is a 4-digit number.
But take 5? It is not a 2-digit number.
Now take 10 to 99:
- 103 = 1000
- 113 = 1331
No 2-digit number has cube as a 3-digit number.
Therefore, the statement is False.
(iv) The cube of a 2-digit number may have seven or more digits.
Greatest 2-digit number = 99
993 = 970299
970299 has 6 digits, not 7.
Therefore, the statement is False.
(v) Cube numbers have an odd number of factors.
Only perfect squares have an odd number of factors.
A cube number does not always have an odd number of factors.
Example: 8 = 23
Factors of 8 are 1, 2, 4, 8 → total 4 factors, which is even.
Therefore, the statement is False.
4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
(i) 1331
We know 113 = 1331
So, ∛1331 = 11.
(ii) 4913
We know 173 = 4913
So, ∛4913 = 17.
(iii) 12167
We know 233 = 12167
So, ∛12167 = 23.
(iv) 32768
We know 323 = 32768
So, ∛32768 = 32.
5. Which of the following is the greatest? Explain your reasoning.
(i) 673 − 663
(ii) 433 − 423
(iii) 672 − 662
(iv) 432 − 422
Use identities:
For squares:
a2 − b2 = (a − b)(a + b)
For cubes:
a3 − b3 = (a − b)(a2 + ab + b2)
(i) 673 − 663
= (67 − 66)(672 + 67 × 66 + 662)
= 1 × (4489 + 4422 + 4356)
= 13267
(ii) 433 − 423
= (43 − 42)(432 + 43 × 42 + 422)
= 1 × (1849 + 1806 + 1764)
= 5419
(iii) 672 − 662
= (67 − 66)(67 + 66)
= 1 × 133
= 133
(iv) 432 − 422
= (43 − 42)(43 + 42)
= 1 × 85
= 85
Comparing all values:
- (i) = 13267
- (ii) = 5419
- (iii) = 133
- (iv) = 85
Therefore, the greatest is 673 − 663.