$x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$? 
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?

I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit.  Are there some good shortcuts?
This is not homework; I'm preparing for an examination.

Another way:
Since $x= 3^\frac{1}{3} + 3^\frac{-1}{3}$, 
$$ \Rightarrow  x - 3^\frac{-1}{3} = 3^\frac{1}{3}$$
Take cubes on both sides
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$
$$x^3 - \frac{1}{3} -3x = 3$$
Multiply by 3 to get your answer :)
