2a3 + 16b3 – 5a – 10b
Above terms can be written as,
2(a3 + 8b3) – 5(a + 2b)
2(a3 + (2b)3) – 5(a + 2b)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
2[(a + 2b) (a2 – 2ab + 4b2)] – 5(a + 2b)
(a + 2b) (2a2 – 4ab + 8b2 – 5)
(ii) a3 – (1/a3) – 2a + 2/a
(a3 – (1/a)3) – 2a + 2/a
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – 1/a) – (a2 + (a × 1/a) + (1/a)2] – 2(a – 1/a)
(a – 1/a) (a2 + 1 + 1/a2) – 2(a – 1/a)
(a – 1/a) (a2 + 1 + 1/a2 – 2)
(a – 1/a) (a2 + (1/a2) – 1)
More Solutions:
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