Prove that: (1 / a3) + (1 / b3) + (1 / c3) = (1 / a + 1 / b + 1/c)3 + 3 / abc?

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Shambhavi Shinde

we know that, a3 + b3 + c3 = (a + b + c)3 -- 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = -- c, b + c = -- a, c + a = -- b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 -- 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 -- 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 --...
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we know that, a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = – c, b + c = – a, c + a = – b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 – 3(– c / ab)(– a / bc)(– b / ac) = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2) = (1/a + 1/b + 1/c)3 – 3 / abc. Thus, proved. read less
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