# Trivial: Rationalize fraction with a third-degree root

This is a pretty trivial question. How do you rationalize a function with a denominator that contains a third degree root?

Edit: My expression is $\displaystyle{\frac{1}{\sqrt[3]{2}-1}}$.

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You multiply by a suitable conjugate. If it's just the cubic root by itself, multiply and divide by the square; e.g., $\frac{1}{\sqrt[3]{2}} = \frac{(\sqrt[3]{2})^2}{2} =\frac{\sqrt[3]{4}}{2}$. If it's more complicated, e.g. $\frac{1}{\sqrt[3]{2}-1}$, you use factorizations such as $a^3-b^3 = (a-b)(a^2+ab+b^2)$, just like you use $a^2-b^2=(a-b)(a+b)$ to clear square roots. I suspect that if you post your specific fraction, it will be simple to show you how to do it for that one, while these general remarks may be somewhat mysterious-looking. – Arturo Magidin Apr 4 '11 at 20:13
Could you please be more specific about what "contains" means? Are you talking about something like $\displaystyle{\frac{a}{\sqrt[3]{b}} }$, or $\displaystyle{\frac{a}{b\pm\sqrt[3]{c}}}$, or something else? – Jonas Meyer Apr 4 '11 at 20:15
$(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and $(a^3+b^3)=(a+b)(a^2-ab+b^2)$ – yoyo Apr 4 '11 at 20:22
So maybe you can give now the answer yourself? Try to get rid of the cubic root in the denominator by multiplying numerator and denominator by an appropriate constant using the expression $(a^3-b^3)=(a-b)(a^2+ab+b^2)$. – Fabian Apr 4 '11 at 20:40
It's $\sqrt[3]{2}^2 + \sqrt[3]{2} + 1$. Thanks! – Paul Manta Apr 4 '11 at 20:47

To rationalize when you have a denominator of the form $a^{1/3}-b^{1/3}$ (as you do here, with $a=2$ and $b=1$), use the identity $x^3-y^3 = (x-y)(x^2+xy+y^2)$.
So $$2-1 = \left(\sqrt[3]{2}-\sqrt[3]{1}\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\right) = \left(\sqrt[3]{2} - 1\right)\left(\sqrt[3]{4}+\sqrt[3]{2} + 1\right).$$ So, multiply both numerator and denominator by $\sqrt[3]{4}+\sqrt[3]{2}+1$. You get: \begin{align*} \frac{1}{\sqrt[3]{2}-1} &= \frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}\\ &= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{(\sqrt[3]{2})^3 - 1^3)}\\ &= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{2-1}\\ &= \sqrt[3]{4} + \sqrt[3]{2} + 1. \end{align*}
If you had a denominator of the form $a^{1/3}+b^{1/3}$, you can instead use the identity $(x^3+y^3) = (x+y)(x^2-xy+y^2)$.