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?

go to power of three both side 
$$(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3})^3 \\\xrightarrow[(a+b)^3=a^3+b^3+3ab(a+b)]{} 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} \sqrt[3] {3})=3\\
$$ we can subsitute $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ so we have $$ 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3\\$$simplifying
$$ 3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3-12\\(\sqrt[3]{36- x})\sqrt[3] {3}=-3$$ to the power of three 
$$ 3(36- x)=-27\\36-x=-9\\x=45$$
