Problem finding in simple algebra

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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.