How do I rationalize the follwoing trinomial surd ?
$\displaystyle \sqrt[3]{4} - \sqrt[3]{2} + 1$
Remember that
$\displaystyle \displaystyle \sqrt[3]{4} = \sqrt[3]{2 \times 2} = \sqrt[3]{2^2} = (\sqrt[3]{2})^2$.
So $\displaystyle \displaystyle \sqrt[3]{4} - \sqrt[3]{2} + 1 = (\sqrt[3]{2})^2 - \sqrt[3]{2} + 1$.
This is a quadratic in $\displaystyle \displaystyle \sqrt[3]{2}$.