# Math Contest Problem

• Feb 5th 2011, 02:52 PM
eric1299171
Math 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 PM
dwsmith
Quote:

Originally Posted by eric1299171
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!

A contest you are doing now or an old problem?
• Feb 5th 2011, 02:57 PM
eric1299171
It's a problem from the 2007 Luzerne County Math Contest.
• Feb 5th 2011, 03:01 PM
dwsmith
Quote:

Originally Posted by eric1299171
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!

$\displaystyle x^2\cdot\left[x^2+x^{-2}-3\right]=0\Rightarrow x^4-3x^2+1=0$

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

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

$\displaystyle w=\frac{3\pm\sqrt{5}}{2}=\frac{3\pm \sqrt{5}}{2}\Rightarrow x=\pm\sqrt{w}=\pm\sqrt{\frac{3\pm \sqrt{5}}{2}}$
• Feb 5th 2011, 03:11 PM
eric1299171
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 PM
dwsmith
Quote:

Originally Posted by eric1299171
If you use the quadratic formula, wouldn't you get a 5 under the square root instead of a -7?

Correct, I forgot to square.
• Feb 5th 2011, 03:18 PM
eric1299171
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 PM
LoblawsLawBlog
You can do it without explicitly solving for $x$. We have
$\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2 =5$. Since $x$ is positive, $x+1/x=\sqrt{5}$.

Now expand $\left(x+\dfrac{1}{x}\right)^3$.
• Feb 5th 2011, 07:40 PM
eric1299171
Thank you so much! My math teacher and I were working on this for hours. =)