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?

Oh, mistake $(x^2)^2+(\frac1{x^2})^2=(x^2+\frac1{x^2})^2-2$ !!
en
$$x+\frac1x=2\sqrt3$$
and 
$$x-\frac1x=2\sqrt2$$
so
$$x^2+\frac1{x^2}=(x+\frac1x)^2-2=10$$
and
$$x^2-\frac1{x^2}=(x+\frac1x)(x-\frac1x)=4\sqrt6$$
It follows that
$$x^4-\frac1{x^4}=(x^2+\frac1{x^2})(x^2-\frac1{x^2})=40\sqrt6$$
