Let f: R arrow R be a function such that f(x) + 3f((π)/(2) - x) = x...

Given $f(x) + 3f\left(\dfrac{\pi}{2} - x\right) = \sin x$

Replacing $x$ with $\dfrac{\pi}{2} - x$, we get:

$f\left(\dfrac{\pi}{2} - x\right) + 3f(x) = \sin\left(\dfrac{\pi}{2} - x\right) = \cos x$

Multiplying this equation by $3$ and subtracting the first equation gives:

$9f(x) - f(x) = 3\cos x - \sin x$

$8f(x) = 3\cos x - \sin x$

$f(x) = \dfrac{3\cos x - \sin x}{8}$

The maximum value of $f(x)$ is $\alpha = \dfrac{\sqrt{3^2 + (-1)^2}}{8} = \dfrac{\sqrt{10}}{8}$.

Thus, $\alpha^2 = \dfrac{10}{64} = \dfrac{5}{32}$.

The points of intersection of the curves $g(x) = x^2$ and $h(x) = \beta x^3$ are given by:

$x^2 = \beta x^3 \Rightarrow x^2(1 - \beta x) = 0 \Rightarrow x = 0, x = \dfrac{1}{\beta}$

The area of the region bounded by the curves is:

$\int_{0}^{1/\beta} (x^2 - \beta x^3) dx = \left[ \dfrac{x^3}{3} - \dfrac{\beta x^4}{4} \right]_{0}^{1/\beta}$

$= \dfrac{1}{3\beta^3} - \dfrac{1}{4\beta^3} = \dfrac{1}{12\beta^3}$

Given that the area is $\alpha^2$, we have:

$\dfrac{1}{12\beta^3} = \dfrac{5}{32}$

$\beta^3 = \dfrac{32}{60} = \dfrac{8}{15}$

Therefore, $30\beta^3 = 30 \times \dfrac{8}{15} = 16$.

Answer: $16$