The problem of surds and indices If $$ \frac 12 \times \left(\sqrt[3] a + \sqrt[3] b + \sqrt[3] c\right) = \frac 1{\sqrt[3] 3 - 1}\;\;\;(a>b>c)$$
then what is  $a - 2b - 3c $ ?

Use the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$ with $x = \sqrt[3]{3}$ to get $2 = (\sqrt[3]{3}-1)(\sqrt[3]{9} + \sqrt[3]{3} + 1)$, or
$$\frac{1}{\sqrt[3]{3}-1} = \frac{1}{2}(\sqrt[3]{9}+\sqrt[3]{3} + \sqrt[3]{1})$$
So $a = 9$, $b = 3$, $c = 1$. 
