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?
 A: \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*} 
A: 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 $x=45$.
A: 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, 
$$ x = 45$$
A: Start with your idea $a=6+\sqrt{x}$ and $b=6-\sqrt{x}$. You have $$ab=36-x \text{, } \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{3} \text{ and } a+b=12$$ Raise the middle equation to power $3$ you get $$a+b +3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b})=3$$ and using the initial assumption $$12 +3\sqrt[3]{ab}\sqrt[3]{3}=3 \text{ or } \sqrt[3]{ab}\sqrt[3]{3}=-3$$ and finally $$ab=-9$$ which allows to conclude to $x=45$ as $ab=36-x$.
A: $$6-\sqrt{x} = t^3$$
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 $\varphi = \frac{\sqrt{5}+1}{2}$
A: 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 
$$ 3(36- x)=-27\\36-x=-9\\x=45$$
