Problem finding in simple algebra It is given,
$$x= \sqrt{3}+\sqrt{2}$$
How to find out the value of $$x^4-\frac{1}{x^4}$$/
The answer is given $40 \sqrt{6}$ but my answer was not in a square-root form
I have done in thsi way:
$$x+ \frac{1}{x}= 2 \sqrt{3}$$
Then,
$$(x^2)^2-\left(\frac{1}{x^2}\right)^2= \left(x^2 + \frac{1}{x^2}\right)^2-2$$
But this way is not working. Where I am wrong?

Note that $\cfrac 1x=\cfrac 1{\sqrt 3+\sqrt 2}=\cfrac {\sqrt 3-\sqrt 2}{\sqrt 3-\sqrt 2}\cdot\cfrac 1{\sqrt 3+\sqrt 2}=\sqrt 3-\sqrt 2$
So you need to find $(\sqrt 3+\sqrt 2)^4-(\sqrt 3-\sqrt 2)^4$
Now note that the terms which have even powers of $\sqrt 2$ will cancel, and the odd powers will double - so we are left with the two terms in the binomial expansion with coefficient $\binom 41=\binom 43=4$ so we get $$2\cdot\binom 41\left((\sqrt3)^3\sqrt2+\sqrt 3(\sqrt 2)^3\right)=2\cdot4\cdot\sqrt 6\cdot(3+2)=40\sqrt 6$$
I've done this longhand, to show the detail.
