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

`\frac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=(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

`(2x^2(1-x))/((1+x+x^2)(1-x))=(2x^2(1-x))/(1-x^3)`

so that

`\frac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=\frac{2\sqrt[3]{9}(1-\sqrt[3]{3})}{1-3}=\sqrt[3]{9}(\sqrt[3]{3}-1)=3-\sqrt[3]{9}`