Page 68 - Maths Skills - 8

66 Maths

Let’s Attempt

Example 1: Find the cube root of 1000.
Solution: (i) Factorising 1000 into its prime factors, we get 2 1000

1000 = 2 × 2 × 2 × 5 × 5 × 5 2 500
(ii) Arranging them into groups taking three equal factors at a time. 2 250
5
125
1000 = 2 × 2 × 2 × 5 × 5 × 5 5 25
(iii) Choose one factor from each group and multiply them. 5 5
i.e., 2 × 5 = 10 1
Thus, 10 is the cube root of 1000.
2 216
Example 2: Find the cube root of – 216. 2 108

Solution: We have, 216 216 2 54
3
3
3 27
Now, factorising 216 into its prime factors, we get 216 = 2 × 2 × 2 × 3 × 3 × 3 3 9
Now, 3 216 3 3
23 6
1
3
∴ 3 − 216 = − 216 = − 6
2 512
Example 3: Find the cube root of – 512 × – 1000. 2 256 2 1000

Solution: We know that, 3 ( 512 ) ( 1000 ) 2 128 2 500
64
2
2
250
3
3
512 1000 512 1000 2 32 5 125
3
3
Now, 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 2 16 5 25
2 8 5 5
So 3 512 = 2 × 2 × 2 = 8 2 4 1
And 1000 = 2 × 2 × 2 × 5 × 5 × 5 2 2
So 3 1000 1
2510
=
3
3
∴ 3 (− 512 ) × (−1000 ) = − 512 × − 1000 = −× −8 1080 [ (–) × (–) = + ]
Example 4: Prove that: 216 27 3 216 3 27. 2 216 3 27
3
2 108 3 9
Solution: LHS = 216 27× 2 54 3 3
3
3 27 1
3
= 2 ×× ×××××× = ××=2 2 333333 23 318
3 9
3
3
RHS 216 27 3 3
1
× ×= 23(
= 2 ×× ××× ×2 2 333 3 333 × ) × =318
3
Hence, LHS = RHS