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Question

If a+2b+3c=0,prove that a3+8b3+27c3=18abc

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Solution

A+2b+3c=0 then

a+2b=-3c

(a+2b)^3=(-3c)^3 (By cubing)

a^3+(2b)^3+3(a)(2b)(a+2b) = -27c^3

a^3+8b^3+6ab (-3c) = -27c^3 [a+2b=-3c]

a^3+8b^3-18abc = -27c^3

a^3+8b^3+27c^3 = 18abc

Hence proved...

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