Given that $x$ is real and $x^3+\frac{1}{x^3}=52$, find $x+\frac{1}{x}$.
tysm!!!
+600 $\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}=52+3\left(x+\frac{1}{x}\right)$
$x+\frac{1}{x}=k\to k^3=52+3k$. We can guess and check or use rational root theorem to find $k=4\implies\boxed{x+\frac{1}{x}=4}$.
