<s>[INST] 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.
 [/INST] We have this:


$$(x + 9)^{\frac{1}{3}} - (x-9)^{\frac{1}{3}} = 3$$


Then:
$$(x + 9)^{\frac{1}{3}} -3 = (x-9)^{\frac{1}{3}}$$
$$((x + 9)^{\frac{1}{3}} -3)^3 = ((x-9)^{\frac{1}{3}})^3$$
$$x + 9 -9(x+9)^{\frac{2}{3}} + 27(x+9)^{\frac{1}{3}} -27 = x -9$$
$$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =-9-9 +27 $$
$$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =9 $$
$$3(x+9)^{\frac{1}{3}} - (x+9)^{\frac{2}{3}} = 1$$
$$(x+9)^{\frac{2}{3}} - 3(x+9)^{\frac{1}{3}} + 1 = 0$$
Let $y = (x+9)^{\frac{1}{3}}$
Then we have:
$$y^2 -3y +1 - 0$$
$$y = \frac{3 \pm \sqrt{5}}{2} = (x +9)^{\frac{1}{3}}$$
$$(\frac{72+32\sqrt{5}}{8})^{\frac{1}{3}} = (x + 9)^{\frac{1}{3}}$$
$$\frac{72+32\sqrt{5}}{8} = x + 9$$
$$72 + 32\sqrt{5} = 8x + 72$$
$$8x = 32\sqrt{5}$$
Hence:


$$x = 4\sqrt{5} = \sqrt{80}$$



So $x^2$ would be $80$
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