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?

\begin{align*} a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
6+\sqrt{x} + 6-\sqrt{x} + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
12 + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= -9 \end{align*} 
Divide by $3\sqrt[3]{ab}$:
$$\sqrt[3]{b} + \sqrt[3]{a} = \frac{-9}{3\sqrt[3]{ab}}$$
Using the original equation we see that the right handed side equals $\sqrt[3]{3}$. We use that. Then substituting $a=6+\sqrt{x}$, $b=6-\sqrt{x}$ will yield a simple lineair equation in $x$ which we can solve. 
\begin{align*} \frac{-9}{3\sqrt[3]{ab}} &= \sqrt[3]{3} \\
\frac{-729}{27ab}&=3 \\
ab &=-9 \\ \\ (6+\sqrt{x})(6-\sqrt{x}) &=-9  \\
36-x &= -9 \\
x &= 45
\end{align*} 
