Let $$f : \mathbb{R} \to \mathbb{R}$$ be a function such that $$f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x, x \in R$$.Let  the maximum value of $$f$$ on $$\mathbb{R}$$ be  $$\alpha$$. The area of the region bounded by the curves $$g(x) = x^2$$ and $$h(x) = \beta x^3$$, $$\beta > 0$$, is $$\alpha^2$$. Then $$30\beta^3$$ is equal to :

The functional equation is
$$f(x) + 3f\!\left(\frac{\pi}{2}-x\right) = \sin x \qquad -(1)$$

Replace $$x$$ by $$\frac{\pi}{2}-x$$ in $$-(1)$$ to obtain
$$f\!\left(\frac{\pi}{2}-x\right) + 3f(x) = \sin\!\left(\frac{\pi}{2}-x\right) = \cos x \qquad -(2)$$

Let $$A = f(x)$$ and $$B = f\!\left(\frac{\pi}{2}-x\right)$$. Then $$-(1)$$ and $$-(2)$$ become
$$A + 3B = \sin x \qquad -(3)$$ $$3A + B = \cos x \qquad -(4)$$

Solve the simultaneous linear equations. Multiply $$-(3)$$ by $$3$$ and subtract $$-(4)$$:
$$3A + 9B - (3A + B) = 3\sin x - \cos x$$ $$8B = 3\sin x - \cos x \;\;\Longrightarrow\;\; B = \frac{3\sin x - \cos x}{8}$$

Substitute $$B$$ in $$-(3)$$:
$$A = \sin x - 3B = \sin x - 3\!\left(\frac{3\sin x - \cos x}{8}\right) = \frac{-\sin x + 3\cos x}{8}$$ Hence
$$f(x) = \frac{3\cos x - \sin x}{8}$$

The expression $$3\cos x - \sin x$$ is of the form $$P\cos x + Q\sin x$$ whose amplitude is $$\sqrt{P^{2}+Q^{2}} = \sqrt{3^{2}+(-1)^{2}} = \sqrt{10}$$. Therefore the maximum value of $$f$$ is
$$\alpha = \frac{\sqrt{10}}{8}$$

Now consider $$g(x)=x^{2}$$ and $$h(x)=\beta x^{3}\;(\beta \gt 0)$$. Points of intersection satisfy $$x^{2} = \beta x^{3}\;\Longrightarrow\; x = 0$$ or $$x=\frac{1}{\beta}$$.

For $$0 \lt x \lt \frac{1}{\beta}$$,\; $$x^{2} \gt \beta x^{3}$$, so the required area is
$$A = \int_{0}^{1/\beta} \bigl(x^{2} - \beta x^{3}\bigr)\,dx$$

Integrate term by term:
$$A = \left[\frac{x^{3}}{3} - \frac{\beta x^{4}}{4}\right]_{0}^{1/\beta} = \frac{1}{3\beta^{3}} - \frac{1}{4\beta^{3}} = \frac{1}{12\beta^{3}}$$

Given that this area equals $$\alpha^{2}$$,
$$\frac{1}{12\beta^{3}} = \left(\frac{\sqrt{10}}{8}\right)^{2} = \frac{10}{64} = \frac{5}{32}$$

Solve for $$\beta^{3}$$:
$$\beta^{3} = \frac{1}{12}\,\Big/\!\frac{5}{32} = \frac{32}{60} = \frac{8}{15}$$

Finally,
$$30\beta^{3} = 30 \times \frac{8}{15} = 16$$

Hence, the required value is 16.