If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

Question

Arnab · Answer

a + b + c = 0 (given) or, b + c = - a or, c + a = - b or, a + b = -c (b+c)&sup2;/3bc + (c+a)&sup2;/3ac + (a+b)&sup2;/3ab = (-a)&sup2;/3bc + (-b)&sup2;/3ac + (-c)&sup2;/3ab = (a)&sup2;/3bc + (b)&sup2;/3ac + (c)&sup2;/3ab = a&sup3;+b&sup3;+c&sup3;/3abc = 3abc/3abc ( a&sup3;+b&sup3;+c&sup3; = 3abc) = 1 hence proved.

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If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

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If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

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