SOLUTION: Please, help me in this question!! I need it very urgently.. Please Please... If {(a^3)+(b^3)+(c^3)}=3abc and a+b+c=0, show that [{(b+c)^2}/3bc]+ [{(c+a)^2}/3ac]+ [{(a+b)^2}/3ab

Question 757057: Please, help me in this question!! I need it very urgently.. Please Please...
If {(a^3)+(b^3)+(c^3)}=3abc and a+b+c=0, show that [{(b+c)^2}/3bc]+ [{(c+a)^2}/3ac]+ [{(a+b)^2}/3ab]=1.

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Divide the first equation by 3abc. This gives a^2/3bc + b^2/3ac + c^2/3ab = 1
Since a + b + c = 1, a = -(b+c) -> a^2 = (b+c)^2
And so on for the numerators b^2 and c^2.
Making the substitutions gives (b+c)^2/3bc + (c+a)^2/3ac + (a+b)^2/3ab
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