If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$ 
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$

This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.

Let $\,a = \sqrt[3]3.\, $  $\, (\!\overbrace{x\!-\!3}^{\Large a^2+a}\!)^2 = \overbrace{a^4}^{\Large 3a}\!+\overbrace{2a^3}^{\Large 6}+a^2\,$   
Therefore  $\ \color{#0a0}{x^2\!-6x\!+\!3\, =\, a^2\!+\!3a}.\ $ By very simple arithmetic we have:
$\begin{align}  f(x)\, & =\ (x-3)\,(\color{#0a0}{x^2\!-6x+3})\, -\, 3(x+1)\ \ {\rm\ via\ \ Division\ Algorithm}\\
 & = (a^2\!+\!a)\,\color{#0a0}{(a^2\!+\!3a)}\, -\, 3(a^2+a+4)\\
 &= \underbrace{a^4}_{\Large 3a}\!\!+\underbrace{4a^3}_{\Large\! 4\,\cdot\, 3}\!+3a^2\, -\, 3(a^2+a+4)\, =\, 0
\end{align}$
