<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] This method doesn't use the $AM-GM$ inequality.
$a^3 +b^3 +c^3 = (a+b+c)^3 -3(a+b+c)(ab+ac+bc)+3abc$
Applying it to your question, we have 
$\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ 
= $(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})^3 -3(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})(\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z}) +3$
=$(7^3)-3(7)(9) +3$
=$157$
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