Radical‑Exponent Manipulation / “Sum‑of‑Cubes” Trick

Rapid Solution (under 30 s)

Set $a=\sqrt[3]{2}$ and $b=\sqrt[3]{4}= \sqrt[3]{2^{2}}$.

Then $x=a+b$.

Use the cube expansion

$$ \begin{aligned} x^{3} &= a^{3}+b^{3}+3ab(a+b) \ &= 2 + 4 + 3\bigl(a,b\bigr),x. \end{aligned} $$

But $ab=\sqrt[3]{2},\sqrt[3]{4}=\sqrt[3]{8}=2$.

So

$$ x^{3}=6 + 3(2)x = 6 + 6x. $$

Re‑arrange:

$$ x^{3}-6x = \boxed{6} $$