15. Given that $\sqrt[3]{17-\frac{27}{4} \sqrt{6}}$ and $\sqrt[3]{17+\frac{27}{4} \sqrt{6}}$ are the roots of the equation
$$
x^{2}-a x+b=0 \text {, }
$$
find the value of $a b$.
15. Answer: 10 .
Let $x_{1}=\sqrt[3]{17-\frac{27}{4} \sqrt{6}}$ and $x_{2}=\sqrt[3]{17+\frac{27}{4} \sqrt{6}}$. Then
$$
\begin{aligned}
b & =x_{1} x_{2}=\sqrt[3]{\left(17-\frac{27}{4} \sqrt{6}\right)\left(17+\frac{27}{4} \sqrt{6}\right)}=\sqrt[3]{\frac{125}{8}}=\frac{5}{2}, \\
a^{3} & =\left(x_{1}+x_{2}\right)^{3}=x_{1}^{3}+x_{2}^{3}+3 x_{1} x_{2}\left(x_{1}+x_{2}\right) \\
& =\left(17-\frac{27}{4} \sqrt{6}\right)+\left(17+\frac{27}{4} \sqrt{6}\right)+3 a b=34+\frac{15}{2} a .
\end{aligned}
$$

Then $a=4$ and thus $a b=10$.