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.

If $(9+x)^\frac 13 + (9-x)^\frac 13 = 3$,
then there is a $y$ such that $(9+x)^\frac 13 = (\frac 32+y)$ and $(9-x)^\frac 13 = (\frac 32-y)$.
Taking cubes of both equations you get $9 \pm x = (\frac {27} 8+\frac 92y^2) \pm (\frac {27} 4 +y^2)y$,
and so $9 = \frac {27} 8+\frac 92y^2$ and $x = (\frac {27} 4 +y^2)y$.
This gives you $y^2 = \frac 54$, then $x = \frac {32} 4 y = 8y$, and so $x^2 = 64y^2 = 16 \times 5 = 80$.
So $(9 + \sqrt {80})^\frac 13 + (9 - \sqrt {80})^\frac 13 = (\frac {3 + \sqrt 5}2) + (\frac {3 - \sqrt 5}2) = 3$ .
As a bonus, you also know the values of the individual cube roots.
