ใช้ความจริงที่ว่า $\arcsin{x}+\arccos{x}=\frac{\pi}{2}$ ครับ

$\arcsin^3{x}+\arccos^3{x}=(\arcsin{x}+\arccos{x})(\arcsin^2{x}-\arcsin{x}\arccos{x}+\arccos^2{x})$
$=(\arcsin{x}+\arccos{x})((\arcsin{x}+\arccos{x})^2-3\arcsin{x}\arccos{x})$
แต่ $\arcsin{x}+\arccos{x}=\frac{\pi}{2}$
$\therefore\arcsin^3{x}+\arccos^3{x}=\frac{\pi}{2}\left(\frac{\pi^2}{4}-3\arcsin{x}\arccos{x}\right)$
$=\frac{\pi^3}{8}-\frac{3\pi}{2}\arcsin{x}\arccos{x}$
$=\frac{\pi^3}{8}-\frac{3\pi}{2}\arcsin{x}(\frac{\pi}{2}-\arcsin{x})$
$=\frac{\pi^3}{8}-\frac{3\pi^2}{4}\arcsin{x}+\frac{3\pi}{2}\arcsin^2{x}$
$=\frac{3\pi}{2}\left(\arcsin{x}-\frac{\pi}{4}\right)^2+\frac{\pi^3}{32}$
แต่ $-\frac{\pi}{2}\leq\arcsin{x}\leq\frac{\pi}{2}$
$-\frac{3\pi}{4}\leq\arcsin{x}-\frac{\pi}{4}\leq\frac{\pi}{4}$
$0\leq(\arcsin{x}-\frac{\pi}{4})^2\leq\frac{9\pi^2}{16}$
$\therefore a=\frac{\pi^3}{32}\leq\frac{3\pi}{2}\left(\arcsin{x}-\frac{\pi}{4}\right)^2+\frac{\pi^3}{32}\leq\frac{7\pi^3}{8}=b$

$\therefore\displaystyle\frac{b}{a}=\frac{\frac{7\pi^3}{8}}{\frac{\pi^3}{32}}=28$ Ans.