Question -
Answer -
We know that (a + bf = a3 + b3 +3ab(a + b) and (a – b)3= a3 – b3 – 3ab(a – b)
Therefore,
(i) (103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 x 100 x 3(100 +3) {∵ (a +b)3 = a3 + b3 + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)3 = (100 – 2)3
= (100)3 – (2)3 – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)3 = (600 – 2)3
= (600)3 – (2)3 – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299