GIF of root

tha $f(x) = x^5 - x^3 +x - 2$ is such that $f'(x) = 5x^4 - 3x^2 + 1 = \big(1 - \tfrac32x^2\big)^2 + \tfrac{11}{4}x^4 > 0$ for all $x$ . Thus $f$ is increasing, and so there is a unique real root $\alpha$ of $f$ . Since $f(1) = -1 < 0 < 24 = f(2)$ , we know that $1 < \alpha < 2$ . Now $\begin{aligned} \alpha^6 & = \; \alpha \times \alpha^5 \; = \; \alpha(\alpha^3 - \alpha + 2) \; =\; \frac{\alpha^5 - \alpha^3 + 2\alpha^2}{\alpha} \; = \; \frac{2 - \alpha + 2\alpha^2}{\alpha} \\ & = \; 2\big(\alpha + \alpha^{-1}\big) - 1 \end{aligned}$ Now $x \mapsto x + x^{-1}$ is increasing for $x > 1$ . Since $1 < \alpha < 2$ we deduce that $2 < \alpha + \alpha^{-1} < \tfrac52$ , and hence $3 < \alpha^6 < 4$ . Hence $\lfloor \alpha^6 \rfloor = \boxed{3}$ .