Let {x} denote the fractional part of x and f(x)=(cos^-1(1-{x}^2) sin^-1(1-{x}))/({x}-{x}^3), x ≠ 0. If L and R respectively denotes the left hand…

Let $\displaystyle \{x\}$ denote the fractional part of $\displaystyle x$ and $\displaystyle f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If L and R respectively denotes the left hand limit and the right hand limit of $\displaystyle f(x)$ at $\displaystyle x=0$, then $\displaystyle \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)$ is equal to $\displaystyle \_\_\_\_$.