# Math Help - Math Contest Problem

1. ## 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!

2. 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?

3. It's a problem from the 2007 Luzerne County Math Contest.

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

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

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

$\displaystyle w=\frac{3\pm\sqrt{5}}{2}=\frac{3\pm \sqrt{5}}{2}\Rightarrow x=\pm\sqrt{w}=\pm\sqrt{\frac{3\pm \sqrt{5}}{2}}$

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

6. 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.

7. 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.

8. 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 $\left(x+\dfrac{1}{x}\right)^3$.

9. Thank you so much! My math teacher and I were working on this for hours. =)