Simplify the denominator

Question

having difficulties getting this..How to simplify by rationalizing the denominator?

$(\frac{\sqrt2+1}{\sqrt2-1})^2-(\frac{\sqrt2-1}{\sqrt2+1})^2$

Answer 1

Hint: It is $$\frac{\sqrt{2}+1}{\sqrt{2}-1}=(\sqrt{2}+1)^2$$ and $$\frac{\sqrt{2}-1}{\sqrt{2}+1}=(\sqrt{2}-1)^2$$ and we get $$(\sqrt{2}+1)^4-(\sqrt{2}-1)^4=((\sqrt{2}+1)^2-(\sqrt{2}-1)^2)((\sqrt{2}+1)^2+(\sqrt{2}-1)^2)=(\sqrt{2}+1+\sqrt{2}-1)(\sqrt{2}+1-\sqrt{2}+1)(2+1+2\sqrt{2}+2+1-2\sqrt{2})=4\sqrt{2}(2+1+2+1)=24\sqrt{2}$$

Answer 2

If $a=\sqrt2+1,b=\sqrt2-1,ab=1,a-b=2,a+b=?$

$$\left(\dfrac ab\right)^2-\left(\dfrac ba\right)^2=\dfrac{a^4-b^4}{(ab)^2}=(a^2+b^2)(a+b)(a-b)$$

$a^2+b^2=(a-b)^2+2ab=?$