Question

Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

Answer 1

Let $x=\sqrt[3]{3}$ and $x^2=\sqrt[3]{9}$. Then

$\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=\dfrac{2x^2}{1+x+x^2}$

Multiply the numerator and denominator by $1-x$. The motivation for this is the rule for factoring a difference of cubes:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Doing so gives

$\dfrac{2x^2(1-x)}{(1+x+x^2)(1-x)}=\dfrac{2x^2(1-x)}{1-x^3}$

so that

$\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=\dfrac{2\sqrt[3]{9}(1-\sqrt[3]{3})}{1-3}=\sqrt[3]{9}(\sqrt[3]{3}-1)=3-\sqrt[3]{9}$