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?
 A: 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.
A: 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
$$x^4-\frac1{x^4}=(x^2+\frac1{x^2})(x^2-\frac1{x^2})=40\sqrt6$$
A: 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.
