# Rationalization of denominator with nth roots

I would need your help to solve a problem with the rationalization:

$$\sqrt{a}-\sqrt{b}\over \sqrt[4]a+\sqrt[4]b$$

I think I'm doing something wrong with the products and signs of the fourth roots.

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@monhawk,@yiyuan lee:thank you both for your help – Mik23 Apr 14 '14 at 10:49
Note that whenever you have square roots in the numerator or denominator,always think about multiplying by the conjugate,using the difference of squares formula or both. – rah4927 Apr 14 '14 at 11:39

## 3 Answers

Hint: Assume that $\sqrt{h} = (\sqrt[4]{h})^2$ and the formulae for the difference of squares.

$$\frac{\sqrt[4]{a}^2 - \sqrt[4]{b}^2}{\sqrt[4]{a} + \sqrt[4]{b}} = \frac{(\sqrt[4]{a} - \sqrt[4]{b})(\sqrt[4]{a} + \sqrt[4]{b})}{\sqrt[4]{a} + \sqrt[4]{b}}$$ And finally we get: $$\sqrt[4]{a} - \sqrt[4]{b}$$

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We have:

\begin{align}\frac{\sqrt{a}-\sqrt{b}}{ \sqrt[4]a+\sqrt[4]b} &= \frac{\sqrt{a}-\sqrt{b}}{ \sqrt[4]a+\sqrt[4]b}\cdot\frac{\sqrt[4]a-\sqrt[4]b}{\sqrt[4]a-\sqrt[4]b}\\ &=\frac{(\sqrt{a}-\sqrt{b})(\sqrt[4]a-\sqrt[4]b)}{\sqrt{a} - \sqrt{b}}\\ &= \sqrt[4]a-\sqrt[4]b\end{align}

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Write $x=\sqrt[4]a$ and $y=\sqrt[4]b$ to avoid tired fingers; you are looking at $\frac{x^2-y^2}{x+y}$ and want to get rid of the denominator, or at least write in in terms of $a=x^4$ and $b=y^4$. Here the simplest way is to recognise that the numerator factors as $x^2-y^2=(x+y)(x-y)$ so the result is simply $x-y=\sqrt[4]a-\sqrt[4]b$. End of problem.

For a less opportunistic approach (the numerator won't always be so kind as to contain the denominator as factor), you may also attain your goal by multiplying numerator and denominator by $x^3-x^2y+xy^2-y^3$, which makes the denominator $x^4-y^4=a-b$.

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