Question

Express with rational denominator:
$\frac{\sqrt[3]{33}-1}{\sqrt[3]{33}+1}$

Answer 1

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

Here we have to rationlize $\sqrt[3]{33}+1$, so we have to convert it into $a^3+b^3$,where $a=\sqrt[3]{33},b=1$ form to do it.

Hence, multiplying the numerator and denominator with $a^2-ab+b^2$ which is $(\sqrt[3]{33}{)}^2-\sqrt[3]{33}+1$.

We get $\frac{(\sqrt[3]{33}-1)\left(\right(\sqrt[3]{33}{)}^2-\sqrt[3]{33}+1)}{(\sqrt[3]{33}+1)\left(\right(\sqrt[3]{33}{)}^2-\sqrt[3]{33}+1)}$

$=\frac{2-2(\sqrt[3]{33}{)}^2+2\sqrt[3]{33}}{4}$

$=\frac{1-(\sqrt[3]{33}{)}^2+\sqrt[3]{33}}{2}$