Given that $x$ is real and $x^3+\frac{1}{x^3}=52$, find $x+\frac{1}{x}$.
Specifically, I recall that $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$. If I isolate $x^3+\frac{1}{x^3}$ on one side, I get $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$. This means that I can substitute $52$ for $x^3+\frac{1}{x^3}$ in the given equation and get $52=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$. To simplify this equation, I let $y=x+\frac{1}{x}$ and rewrite it as $y^3-3y-52=0$. This is a cubic equation in $y$, which I can try to factor by looking for rational roots. By the rational root theorem, the possible rational roots of this equation are $\pm 1, \pm 2, \pm 4, \pm 13, \pm 26, \pm 52$. I find that $y=4$ is a root, since $4^3-3(4)-52=0$. This means that I can factor out $(y-4)$ from the equation and get $(y-4)(y^2+4y+13)=0$. The other factor, $y^2+4y+13$, has no real roots, since its discriminant is negative. Therefore, the only real solution for $y$ is $y=4$. But $y$ is the same as $x+\frac{1}{x}$, so I have found the answer. 