<s>[INST] find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ 
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$

let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
 [/INST] Start with your idea $a=6+\sqrt{x}$ and $b=6-\sqrt{x}$. You have $$ab=36-x \text{, } \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{3} \text{ and } a+b=12$$ Raise the middle equation to power $3$ you get $$a+b +3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b})=3$$ and using the initial assumption $$12 +3\sqrt[3]{ab}\sqrt[3]{3}=3 \text{ or } \sqrt[3]{ab}\sqrt[3]{3}=-3$$ and finally $$ab=-9$$ which allows to conclude to $x=45$ as $ab=36-x$.
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