If the area enclosed by the parabola ${\text{P}}_1:\mathrm{\hspace{0.17em}}2\mathrm{y}=5{\mathrm{x}}^2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\mathrm{P}}_2:{\mathrm{x}}^2-\mathrm{y}+6=0$  is equal to the area enclosed by ${\text{P}}_1\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\mathrm{\alpha x},\hspace{0.17em}\hspace{0.17em}\mathrm{\alpha }>0,\hspace{0.17em}\hspace{0.17em}\mathrm{then}\hspace{0.17em}\hspace{0.17em}{\mathrm{\alpha }}^3$  is equal to …… 

$$\begin{gathered} x^2+6=\frac{5}{2}{x}^2\Rightarrow x=\pm 2 \\ \text{ Area between }{P}_1\text{ and }{P}_2 \\ \begin{array}{l}{A}_1={\int }_{-2}^2 \left(x^2+6\right)-\frac{5}{2}{x}^{2}dx\\ =2{\int }_0^2 \left(6-\frac{3}{2}{x}^2\right)dx=2{\left[6x-\frac{x^3}{2}\right]}_0^2=16\end{array} \end{gathered}$$
$$\begin{gathered} \text{ Area between }{P}_1\text{ and }y=\alpha x \\ {A}_2=\underset{0}{\overset{\frac{2\mathrm{\alpha }}{5}}{\int }}(\mathrm{\alpha x}-\frac{5{\mathrm{x}}^2}{2})\mathrm{dx} \\ {\left.=\frac{\alpha x^2}{2}-\frac{5}{6}{x}^3\right]}_0^{\frac{2\alpha }{5}}=\frac{2{\alpha }^3}{75} \end{gathered}$$
$\begin{array}{l}{A}_1=A_2\Rightarrow \frac{2{\alpha }^3}{75}=16\\ {\alpha }^3=600\end{array}$