Rationalize this I'm having trouble rationalizing the denominator of this fraction. Would you kindly explain this to a fellow self-learning math student?
$$\frac{10}{\sqrt[4]{3}-1}$$  knowing that $a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}\cdot b+a^{n-3}\cdot b^2+...+a\cdot b^{n-2}+b^{n-1}\right)$
$a,b\in \mathbb{R}$ and $n\in \mathbb{N}$

Couldn't see how the tip helps.
$$
\frac{10}{\sqrt[4]{3}-1} = \frac{10}{\sqrt[4]{3}-1}\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1} = \frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}=\frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{10(\sqrt[4]{3}+1)(\sqrt{3}+1)}{2}
$$
