Rationalize this

Question

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 $n\in \mathbb{N}$

Answer 1

Couldn't see how the tip helps.

$$ \frac{10}{\sqrt[4]{3}-1} = \frac{10}{\sqrt[4]{3}-1}\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1} = \frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}=\frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{10(\sqrt[4]{3}+1)(\sqrt{3}+1)}{2} $$

Answer 2

$$\dfrac{10}{\sqrt[4]{3}-1} = \dfrac{5\cdot (\sqrt[4]{3^4}-1^4)}{\sqrt[4]{3}-1} = 5\left(\sqrt[4]{3^3}\cdot 1^0+\sqrt[4]{3^2}\cdot 1^1+\sqrt[4]{3^1}\cdot 1^2+\sqrt[4]{3^0}\cdot 1^3\right)$$