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?

$$6-\sqrt{x} = t^3$$
then
$$6+\sqrt{x} = t^3+2\sqrt{x}$$
The expression becomes:
$$t + \sqrt[3]{t^3+2\sqrt{x}} = \sqrt[3]{3}$$
Substitute back $\sqrt{x}$ in terms of t, rewrite so the cube root is alone. Raise to 3, solve 3rd degree polynomial equation. Test all roots.

Fun fact: The valid $t$ solution happens to become closely related to the golden ratio: $t =\sqrt[3]{3}\left(\frac{1+\sqrt{5}}{2}\right) = \sqrt[3]{3}\varphi$, where $\varphi = \frac{\sqrt{5}+1}{2}$
