value of $x^3 - 6x^2 + 6x$ when $x$ is an expression involving radicals

Question

If $x = 2 + 2^{\frac13} + 2^{\frac23}$, then what is the value of $x^3 - 6x^2 + 6x$?

How would I solve it? Surely plugging in is one way, but there's got to be some other way.

Answer 1

$x=2+2^{\frac13}+2^{\frac23}\implies (x-2)^3=(2^{\frac13}+2^{\frac23})^3$

$\implies x^3-8-6x^2+12x=2+4+6(x-2)=6x-6$

$\implies x^3-6x^2+6x=2$

Answer 2

You can factor it as $2(x-1)^3 - x^3 + 2$. Since $(x-1)\left(2^{1/3}-1\right) = 1$ and $x\left(1-2^{-1/3}\right) = 1$ we get $$\left(\frac{2^{1/3}}{2^{1/3}-1}\right)^3 - \left(\frac{1}{1-2^{-1/3}}\right)^3 + 2.$$ So the left two terms cancel out.