If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$ 

If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
    Find $x^2$


I can't seem to solve this question. Any hints or solutions is welcomed.

For $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$ we have an interesting problem.
It's
$$\sqrt[3]{x+9}+\sqrt[3]{9-x}-3=0$$
and since $\sqrt[3]{x+9}=\sqrt[3]{9-x}=-3$ is impossible, it's
$$9+x+9-x-27-9\sqrt[3]{(9+x)(9-x)}=0,$$
which gives $x^2=80$.
I used $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,
where $a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only, which is impossible in our problem.
