<s>[INST] How to relate the $|z^3+z^{-3}|$ with $|z+z^{-1}|$? I got stuck on this seemingly simple question:

If $z$ is a complex number satisfying $|z^3+z^{-3}| \le 2$, then the maximum possible value of $|z+z^{-1}|$ is:
  (A) $2$
  (B) $2^{1/3}$
  (C) $2\sqrt 2$
  (D) $1$

Using the AM-GM inequality, I showed that $|z|^3 + |z|^{-3} \ge 2$ and so I got:
$$ |z^3+z^{-3}| \le 2 \le |z|^3 + |z|^{-3}$$
I don't inderstand how I can remove the cubic power, and bring any of this in terms of $|z|$ and $|z|^{-1}$. What am I missing?
Thank you.
 [/INST] We can write $$z^3+\frac{1}{z^3} = \left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right).$$
Given $$2 \geq \left|z^3+\frac{1}{z^3}\right| = \left|\left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right)\right|\geq \left|z+\frac{1}{z}\right|^3-3\left|z+\frac{1}{z}\right|.$$
Now put $$\left|z+\frac{1}{z}\right|=y.$$
So we get $y^3-3y\leq 2\Rightarrow y^3-3y-2\leq 0\Rightarrow y^3-4y+y-2\leq0$.
So $$\displaystyle y(y-2)(y+2)+1(y-2)\leq 0\Rightarrow (y-2)(y+1)^2\leq0.$$
So we get $$y\leq 2\Rightarrow \left|z+\frac{1}{z}\right|\leq 2$$
because $$(y+1)^2\geq 0.$$
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