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Let $$w=z+\frac{1}{z}.$$Then $$\begin{array}{rl}{w}^3& =z^3+3z^2\cdot \frac{1}{z}+3z\cdot \frac{1}{z^2}+\frac{1}{z^3}\\ & =z^3+\frac{1}{z^3}+3\left(z+\frac{1}{z}\right)\\ & =52+3w,\end{array}$$so $w^3-3w-52=0.$ This equation factors as $(w-4)(w^2+4w+13)=0,$ so $w=4$ or $w^2+4w+13=0.$

For $w^2+4w+13=0,$ completing the square, we get $$(w+2{)}^2=-9.$$Then $w+2=\pm 3i,$ so $w=-2\pm 3i.$

Therefore, the possible values of $w$ are $\boxed{4,-2+3i,-2-3i}.$