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?

Let  $ a = \sqrt[3]{6+\sqrt x} $ and $ b = \sqrt[3]{6-\sqrt x}$, then we have:
$$ \left\{\begin{matrix}
a^3 + b^3 = 12 \\ 
a + b =  \sqrt[3]{3}
\end{matrix}\right. $$
Therefore,
$$ ab = \sqrt[3]{-9}$$
or,
  $$ \sqrt[3]{36 - x} = \sqrt[3]{-9} $$
then, 
$$ x = 45$$
