[Algebra]
Problem: The solutions to the equation $x^{2}-180 x+8=0$ are $r_{1}$ and $r_{2}$. Compute

$$
\frac{r_{1}}{\sqrt[3]{r_{2}}}+\frac{r_{2}}{\sqrt[3]{r_{1}}}
$$

Solution: First note that the solutions of the given equation are real because the equation's discriminant is positive. By Vieta's Formulas, $r_{1}+r_{2}=180(*)$ and $r_{1} r_{2}=8(* *)$. The expression to be computed can be written with a common denominator as

$$
\frac{\sqrt[3]{r_{1}^{4}}+\sqrt[3]{r_{2}^{4}}}{\sqrt[3]{r_{1} r_{2}}}
$$

By $(* *)$, the denominator is equal to $\sqrt[3]{8}=2$. To compute the numerator, first let $S_{k}=\sqrt[3]{r_{1}^{k}}+\sqrt[3]{r_{2}^{k}}$, so that the numerator is $S_{4}$. Then note that

$$
\begin{aligned}
\left(S_{1}\right)^{3} & =r_{1}+3 \sqrt[3]{r_{1}^{2} r_{2}}+3 \sqrt[3]{r_{2}^{2} r_{1}}+r_{2} \\
& =\left(r_{1}+r_{2}\right)+3 \sqrt[3]{r_{1} r_{2}}\left(\sqrt[3]{r_{1}}+\sqrt[3]{r_{2}}\right) \\
& =180+3 \cdot 2 \cdot S_{1}
\end{aligned}
$$

where $(*)$ and $(* *)$ are used to substitute values into the second equality. Next note that $S_{1}^{3}-6 S_{1}-180$ can be factored as $\left(S_{1}-6\right)\left(S_{1}^{2}+6 S_{1}+30\right)$. Because the polynomial $t^{2}+6 t+30$ has no real roots, the unique real solution to $(\dagger)$ is $S_{1}=6$, so $\sqrt[3]{r_{1}}+\sqrt[3]{r_{2}}=6$. Square each side of the previous equation to obtain $S_{2}+2 \sqrt[3]{r_{1} r_{2}}=36$, hence $S_{2}=36-2 \cdot 2$; that is, $\sqrt[3]{r_{1}^{2}}+\sqrt[3]{r_{2}^{2}}=32$. Again, square both sides of this equation to obtain $\sqrt[3]{r_{1}^{4}}+2 \sqrt[3]{r_{1}^{2} r_{2}^{2}}+\sqrt[3]{r_{2}^{4}}=1024$, so $S_{4}+2 \sqrt[3]{r_{1}^{2} r_{2}^{2}}=1024$, from which $S_{4}=1024-2 \cdot 4=1016$. Thus the desired expression equals $\frac{S_{4}}{2}=\frac{1016}{2}=\mathbf{5 0 8}$.

Final answer: 508