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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?

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    $\begingroup$ It should be $-2$ instead of $-4$. $\endgroup$ – Joe Johnson 126 Oct 11 '13 at 16:08
  • $\begingroup$ $x^2=5+2\sqrt{6}$ and $x^4=49+20\sqrt{6}$, so $x^{-4}=(31-20\sqrt{6})/(2401-2400)$. $\endgroup$ – egreg Oct 11 '13 at 16:11
  • $\begingroup$ Er... What's the idea going from the left side of the last equation to the right? It looked like it was heading to a factorization into $(x^2-x^{-2})(x^2+x^{-2})$ but then whatever that is happened... $\endgroup$ – rschwieb Oct 11 '13 at 16:19
  • $\begingroup$ If your main question is "Where am I wrong?", most concisely, you are wrong in the minus sign on the left side of the final equation. It should be plus, which of course does not help you solve the equation. $\endgroup$ – Mark Bailey Oct 11 '13 at 16:21
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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 $$x^4-\frac1{x^4}=(x^2+\frac1{x^2})(x^2-\frac1{x^2})=40\sqrt6$$

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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.

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  • $\begingroup$ I think the point is to be over-fancy. It's from a math class, so the students are learning to find the more elegant solution. $\endgroup$ – Mark Bailey Oct 11 '13 at 16:23
  • $\begingroup$ @mark Strange... why would that be considered "elegant"? This way is computationally simpler apparently, and most people find elegance in simplicity, not overfanciness. Square, square, invert, subtract, done. I would never have thought of using an identity such as that first... $\endgroup$ – rschwieb Oct 11 '13 at 16:29
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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.

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