Could anyone explain this problem to me? If x is a positive real number and x^2 + 1/(x^2) = 3, then x^3 + 1/(x^3) = ? Thanks in advance!

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- Feb 5th 2011, 02:52 PMeric1299171Math Contest Problem
Could anyone explain this problem to me? If x is a positive real number and x^2 + 1/(x^2) = 3, then x^3 + 1/(x^3) = ? Thanks in advance!

- Feb 5th 2011, 02:54 PMdwsmith
- Feb 5th 2011, 02:57 PMeric1299171
It's a problem from the 2007 Luzerne County Math Contest.

- Feb 5th 2011, 03:01 PMdwsmith
$\displaystyle \displaystyle x^2\cdot\left[x^2+x^{-2}-3\right]=0\Rightarrow x^4-3x^2+1=0$

$\displaystyle w^2=x^4\Rightarrow w=x^2$

$\displaystyle w^2-3w+1=0$

$\displaystyle \displaystyle w=\frac{3\pm\sqrt{5}}{2}=\frac{3\pm \sqrt{5}}{2}\Rightarrow x=\pm\sqrt{w}=\pm\sqrt{\frac{3\pm \sqrt{5}}{2}}$ - Feb 5th 2011, 03:11 PMeric1299171
If you use the quadratic formula, wouldn't you get a 5 under the square root instead of a -7?

- Feb 5th 2011, 03:13 PMdwsmith
- Feb 5th 2011, 03:18 PMeric1299171
Thanks. =) Then after you solve for x how do you figure out what x^3 + 1/(x^3) is? I tried plugging back in but it turned really ugly.

- Feb 5th 2011, 05:33 PMLoblawsLawBlog
You can do it without explicitly solving for $\displaystyle x$. We have

$\displaystyle \left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2 =5$. Since $\displaystyle x$ is positive, $\displaystyle x+1/x=\sqrt{5}$.

Now expand $\displaystyle \left(x+\dfrac{1}{x}\right)^3$. - Feb 5th 2011, 07:40 PMeric1299171
Thank you so much! My math teacher and I were working on this for hours. =)