If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
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Is there are typo in the last fraction $\frac{(a+b)^2}{3ab}$? – user103828 Sep 11 '14 at 08:37
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@user103828 Sorry but I can't understand what you want to say.What is typo? Please make it more clear. – SNEHIL SINHA Sep 11 '14 at 08:40
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1in the numerator of your last term it says $(a+b^2)$. Shouldn't it be $(a+b)^2$ like the other terms? – user103828 Sep 11 '14 at 08:48
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@user103828 thanks for pointing it out.It's really a typo! – SNEHIL SINHA Sep 11 '14 at 08:49
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1A side note: $a+b+c=0$ implies $a^3+b^3+c^3=3abc$. – Quang Hoang Sep 11 '14 at 10:40
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$a+b+c=0\iff b+c=-a\implies (b+c)^2=a^2$
$\implies\dfrac{(b+c)^2}{3bc}=\dfrac{a^3}{3abc}$
Actually, $b+c=-a\implies(b+c)^3=(-a)^3$
$\implies -a^3=b^3+c^3+3bc(b+c)=b^3+c^3+3bc(-a)\iff\sum a^3=3abc$
lab bhattacharjee
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$$\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b^2)}{3ab}\\=\frac{a(b+c)^2} {3abc}+\frac{b(c+a)^2}{3abc}+\frac{c(a+b^2)}{3abc}\\=\frac{a^3} {3abc}+\frac{b^3}{3abc}+\frac{c^3}{3abc}\\=\frac{a^3+b^3+c^3}{3abc}\\=\frac{3abc}{3abc}=1$$
creative
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since $$LHS=\dfrac{a^2}{3bc}+\dfrac{b^2}{3ca}+\dfrac{c^2}{3ab}=\dfrac{a^3+b^3+c^3}{3abc}=\dfrac{3abc}{3abc}=1$$
I think this condition $a^3+b^3+c^3=3abc$ is redundant.because $$a+b+c=0\Longrightarrow a^3+b^3+c^3=3abc$$ because $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
math110
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