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# NMTC Problem 5a

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.

Note by Siddharth G
1 year, 11 months ago

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$$x^{2}(y+z)=a^{3}$$ 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. · 1 year, 11 months ago