15. If $\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0$, find the value of $\left\lfloor\alpha^{6}\right\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.
15. Answer: 3
It can be easily seen that the given equation has exactly one real root $\alpha$, since (1) all polynomial equations of degree 5 has at least one real root, and (2) the function $f(x)=$ $x^{5}-x^{3}+x-2$ is strictly increasing since $f^{\prime}(x)=5 x^{4}-3 x^{2}+1>0$ for all real values of $x$. It can also be checked that $\mathrm{f}\left(\frac{1}{2}\right)0$, so that $\frac{1}{2}<\alpha<2$. This is equivalent to $\alpha^{6}<4$ since
$$
\begin{aligned}
\alpha^{6}<4 & \Longleftrightarrow \alpha^{4}-\alpha^{2}+2 \alpha<4 \\
& \Longleftrightarrow \alpha^{5}-\alpha^{3}+2 \alpha^{2}<4 \alpha \\
& \Longleftrightarrow 2 \alpha^{2}-5 \alpha+2<0 \\
& \Longleftrightarrow \frac{1}{2}<\alpha<2
\end{aligned}
$$

In addition, we claim that $\alpha^{6} \geq 3$ since
$$
\begin{aligned}
\alpha^{6} \geq 3 & \Longleftrightarrow \alpha^{4}-\alpha^{2}+2 \alpha \geq 3 \\
& \Longleftrightarrow \alpha^{5}-\alpha^{3}+2 \alpha^{2}-3 \alpha \geq 0 \\
& \Longleftrightarrow 2 \alpha^{2}-4 \alpha+2 \geq 0
\end{aligned}
$$
the last inequality is always true. Hence $3 \leq \alpha^{6}<4$, thereby showing that $\left\lfloor\alpha^{6}\right\rfloor=3$.