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?

The idea you're having to change it to terms of $x^2$ isn't bad, but it seems a little overfancy. (Maybe I overlooked some economy about it, but I haven't seen the benefit yet.)
Why not just calculate it directly? (Hints follow:)
$x^2=3+2+2\sqrt{6}=5+2\sqrt{6}$
$x^4=(5+2\sqrt{6})^2=25+24+20\sqrt{6}=49+20\sqrt{6}$
$\dfrac{1}{x^4}=\dfrac{1}{49+20\sqrt{6}}=\dfrac{49-20\sqrt{6}}{2401-2400}=49-20\sqrt{6}$
You can take it from here, I think.
