<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 $\displaystyle \frac{x}{y} =a \;\;,\frac{y}{z} = b\;\;,\frac{z}{x} = c\;,$ Then $a+b+c=7$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$
and $abc=1$. Then we have to calculate $a^3+b^3+c^3 = $
Now Using $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9\Rightarrow \frac{ab+bc+ca}{abc} = 9\Rightarrow ab+bc+ca = 9$$
Now Let $t=a\;,t=b\;,t=c$ be the roots of cubic equation in terms of $t\;$ Then
Using factor Theorem, We get $(t-a)\;,(t-b)\;,(t-c)$ are the factor of that equation.
So we get $$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$
So we get $$t^3-7t^2+9t-1=0\Rightarrow t^3=7t^2-9t+1$$
Now above we have take $t=a\;,t=b\;,t=c$ be the root of that cubic equation
So we get  $$a^3+b^3+c^3=7(a^2+b^2+c^2)-9(a+b+c)+3 = 7(31)-9(7)+3=157$$
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