If none of \(a,b,c,x,y,z\) is zero and \(\frac { { x }^{ 2 }(y+z) }{ { a }^{ 3 } } =\frac { { y }^{ 2 }(z+x) }{ { b }^{ 3 } } =\frac { { z }^{ 2 }(x+y) }{ { c }^{ 3 } } =\frac { xyz }{ abc } =1\).

Prove that \({a}^3+{b}^3+{c}^3+abc=0\).

This a part of my set NMTC 2nd Level (Junior) held in 2014.

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## Comments

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TopNewest\(x^{2}(y+z)=a^{3}\) and \(xyz=abc\). Dividing the first equation by the second, we get \(x (\frac {1}{y}+\frac {1}{z})=\frac {a^{2}}{bc}\). We can do this two more times with different variables to get two more similar equations. Multiplying all of them,

\(xyz (\frac {1}{y}+\frac {1}{z})(\frac {1}{y}+\frac {1}{x})(\frac {1}{x}+\frac {1}{z})=\frac {(x+y)(y+z)(x+z)}{xyz}=1\) hence

\((x+y)(y+z)(x+z)=xyz\). Expanding and subtracting both sides by \(xyz\), then substituting \(xyz\) with \(abc\), \(x^{2}(y+z)\) with \(a^{3}\) and so on gives the desired result.

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