<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] Step 1 (conjecture): there is some $e>0$ such that
$$
6+\sqrt{x}=(e+\sqrt[3]{3}/2)^3,\quad 6-\sqrt{x}=(-e+\sqrt[3]{3}/2)^3.\tag{$*$}
$$
You can see that if such an $e$ exists then the equality $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ is satisfied. Solving for $e$ is simple:
$$
12=(6+\sqrt{x})+(6-\sqrt{x})=(e+\sqrt[3]{3}/2)^3+(-e+\sqrt[3]{3}/2)^3=\frac{3}{4}+3\sqrt[3]{3}e^2
$$
from which $e=\frac{1}{2}\sqrt[3]{3}\sqrt{5}$.
Step 2 (verify): with $e$ as above, you can verify that ($*$) is satisfied with $\sqrt{x}=3\sqrt{5}$ which is equivalent to $x=45$.
</s>