Rationalization of denominator with nth roots

Question

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

\begin{equation} \sqrt{a}-\sqrt{b}\over \sqrt[4]a+\sqrt[4]b \end{equation}

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

Answer 1

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: $$\sqrt[4]{a} - \sqrt[4]{b}$$

Answer 2

We have:

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

Answer 3

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 $x^4-y^4=a-b$.

Answer 4

If you're dealing with $4$th roots, there's a simple formula $$\frac{1}{\sqrt[4]{a}+\sqrt[4]{b}}=\frac{\left(\sqrt[4]{a}-\sqrt[4]{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}$$ If you're dealing with $n$th roots, you have the sigma notation expressions $$\frac{1}{\sqrt[2n]{a}+\sqrt[2n]{b}}=\frac{\left(\sqrt[2n]{a}-\sqrt[2n]{b}\right)\sum_{k=0}^{n-1}\sqrt[n]{a^{n-k-1}b^{k}}}{a-b}$$ $$\frac{1}{\sqrt[2n+1]{a}+\sqrt[2n+1]{b}}=\frac{\sum_{k=0}^{2n}\sqrt[2n+1]{a^{2n-k}\left(-b\right)^{k}}}{a+b}$$ So you just multiply and simplify to get $$\frac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{a}+\sqrt[4]{b}}=\sqrt[4]{a}-\sqrt[4]{b}$$