$\begin{aligned} & \int_0^\alpha\left(x-x^3\right) d x=\frac{1}{2} \int_0^1\left(x-x^3\right) d x \\ & \left(\frac{x^2}{2}-\frac{x^4}{4}\right)_0^\alpha=\frac{1}{2}\left(\frac{x^2}{2}-\frac{x^4}{4}\right)_0^1 \\ & \left(\frac{2 \alpha^2-\alpha^4}{4}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{4} \cdot \frac{1}{2}\end{aligned}$ $\begin{array}{ll}2 \alpha^2-\alpha^4=\frac{1}{2} & \\ 4 \alpha^2-2 \alpha^4=1 & t=\frac{4 \pm 2 \sqrt{2}}{4} \\ \text { Check 'D' U } & t=1 \pm \frac{1}{\sqrt{2}} \\ \alpha^2=t & \alpha^2=1+\frac{1}{\sqrt{2}}, \quad \alpha^2=1-\frac{1}{\sqrt{2}} \\ 2 t^2-4 t+1=0 & 1>\alpha>\frac{1}{2}\end{array}$