<s>[INST] Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.
 [/INST] let $a= \frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then some manipulations would show that 
$(a+b+c)^3= 3(a+b+c)((a^2+b^2+c^2)) - 2(a^3+b^3+c^3) + 6abc$
we already have $a+b+c = 7, abc=1$, now its left to find what $a^2+b^2+c^2$ is.
$(\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 = (\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 + 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})$
$a^2+b^2+c^2=(\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 = (\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 - 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}) = 7^2-2(9)=37$
so we have 
$7^3 = 3(7)(37)-2(a^3+b^3+c^3)+6$
$((\frac{x}{y})^3+(\frac{y}{z})^3+(\frac{z}{x})^3) = 157$
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