$\therefore$ Area of shaded region

$$={\int }_{-{\left(\frac{1}{2}\right)}^{\frac{3}{2}}}^0{(1-x^{\frac{2}{3}})}^{\frac{3}{2}}+x)dx+{\int }_0^1{(1-x^{\frac{2}{3}})}^{\frac{3}{2}}dx$$

$$={\int }_{-{\left(\frac{1}{2}\right)}^{\frac{3}{2}}}^0{(1-x^{\frac{2}{3}})}^{\frac{3}{2}}dx+{\int }_{-{\left(\frac{1}{2}\right)}^{\frac{3}{2}}}^{0}xdx$$

Let $x={\sin }^3\theta$

$$\therefore dx=3{\sin }^2\theta \cos \theta d\theta$$

$$\begin{array}{rl} & ={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}3{\sin }^2\theta {\cos }^4\theta d\theta +(0-\frac{1}{16})\\ & =\frac{9\pi }{64}+\frac{1}{16}-\frac{1}{16}=\frac{36\pi }{256}=A\\ \therefore & \frac{256A}{\pi }=36\end{array}$$