If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc
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We know that if a + b + c = 0 then a3 + b3 + c3 = 3abc
Given,
a + b + 2c = 0
∴ (a)3 + (b)3 + (2c)3 = 3(a)(b)(2c)
⇒ a3 + b3 + 8c3 = 6abc
Hence Proved.
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