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!
$\displaystyle \displaystyle x^2\cdot\left[x^2+x^{-2}-3\right]=0\Rightarrow x^4-3x^2+1=0$
$\displaystyle w^2=x^4\Rightarrow w=x^2$
$\displaystyle w^2-3w+1=0$
$\displaystyle \displaystyle w=\frac{3\pm\sqrt{5}}{2}=\frac{3\pm \sqrt{5}}{2}\Rightarrow x=\pm\sqrt{w}=\pm\sqrt{\frac{3\pm \sqrt{5}}{2}}$
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 $\displaystyle \left(x+\dfrac{1}{x}\right)^3$.