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.

Let $$\frac{x}{y} = a,\,\,\frac{y}{z} = b,\,\,\frac{z}{x} = c$$ where $$abc = \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1$$
So we have $$\left\{ \begin{gathered}
  a + b + c = 7 \hfill \\
  ab + ac + bc = 9 \hfill \\ 
\end{gathered}  \right.$$
Using the identity $${a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - (ab + ac + bc))$$ which is equivalent to $${a^3} + {b^3} + {c^3} = {(a + b + c)^3} - 3(ab + ac + bc)(a + b + c) + 3abc$$ Now substituting the values we get $${a^3} + {b^3} + {c^3} = {7^3} - 3 \cdot 9 \cdot 7 + 3 = 343 - 189 + 3 = 157$$
