jmc

We know that $${\left(x+\frac{1}{x}\right)}^3=x^3+3(x^2)\left(\frac{1}{x}\right)+3(x){\left(\frac{1}{x}\right)}^2+{\left(\frac{1}{x}\right)}^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right).$$Let $x+\frac{1}{x}=a$. Then our equation is $a^3=x^3+\frac{1}{x^3}+3a$. We know $x^3+\frac{1}{x^3}=52$, so we have $a^3=52+3a$ or $a^3-3a-52=0$. By the rational root theorem, the possible roots of this polynomial equation are the divisors of 52 as well as their negatives: $\pm 1,\pm 2,\pm 4,\pm 13,\pm 26,\pm 52$. Both $\pm 1$ and $\pm 2$ are easy to check by substitution. For $\pm 4$ we can use synthetic division (or substitution), and we find that that $a=4$ is a root. (We could also see this by inspection by writing $a^3-3a=52$ and noting that $4$ works.)

Are there other solutions? Use synthetic division to divide:

\begin{array}{c|cccc} $4%%DISP_2%%amp;$1%%DISP_2%%amp;$0%%DISP_2%%amp;$-3%%DISP_2%%amp;$-52$\\ %%DISP_1%%&$1%%DISP_2%%amp;$4%%DISP_2%%amp;13$&$0$ \end{array} The quotient is $a^2+4a+13$, so $a^3-3a-52=(a-4)(a^2+4a+13)$. The discriminant of $a^2+4a+13$ is $4^2-4(1)(13)=16-52=-36$, which is negative, so there are no other real solutions for $a$. If $x$ is real, $a$ must be real, so we conclude that there are no other values of $x+\frac{1}{x}$. Thus $x+\frac{1}{x}=a=\boxed{4}$.