If $$\sqrt[n]{{abc}} = 1,$$ Prove that $$\frac{\sqrt[n]a}{\sqrt[n]{ab}+\sqrt[n]a+1}+\frac{\sqrt[n]b}{\sqrt[n]{bc}+\sqrt[n]b+1}+\frac{\sqrt[n]c}{\sqrt[n]{ac}+\sqrt[n]c+1}=1.$$
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7First of all note that you can simply ignore all those $\sqrt[n]\cdot$. But why the tag diophantine-equations? – Hagen von Eitzen Dec 07 '13 at 14:10
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Related : http://math.stackexchange.com/questions/569322/simplyfying-the-equation – lab bhattacharjee Dec 07 '13 at 14:45
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Let $x = \sqrt[n]a, y = \sqrt[n]b, z = \sqrt[n]c$. Then you have $xyz = 1$ and $$\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1} = \frac{x}{xy+x+1}+\frac{xy}{1+xy+x}+\frac{1}{x+1+xy}$$
where we multiplied the second term's numerator and denominator by $x$ and the third term's by $xy$.
Macavity
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Yeah, I'm really sorry. I wrote that I was confused, but a mistake in my writing. So I thought to myself this is the result. – mehransalehi Dec 07 '13 at 15:13