Let `{x}` denote the fractional part of x and `f(x) = (cos^(-1) (1-{x}^2) sin^(-1)(1-{x}))/(({x})-{x}^3), x ne 0`. If L and R respectively denotes the left hand limit and the right hand limit of `f(x)` at `x = 0`, then `frac{32}{pi^2} (L^2+R^2)`is equal to ____________.

Answer

To solve the given problem, we need to find the left-hand limit (L) and right-hand limit (R) of the function \( f(x) \) at \( x = 0 \), and then calculate \( \frac{32}{\pi^2}(L^2 + R^2) \).

### Step-by-Step Solution:

1. **Understanding the function**:
The function is defined as:
\[ f(x) = \frac{\cos^{-1}(1 - \{x\}^2) \sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, \quad x \neq 0 \]
where \( \{x\} \) denotes the fractional part of \( x \).

2. **Finding the left-hand limit (L)**:
For \( x \to 0^- \) (approaching from the left), we have:
\[ \{x\} = x + 1 \quad \text{(since \( x \) is negative)} \]
Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = \frac{\cos^{-1}(1 - (x + 1)^2) \sin^{-1}(1 - (x + 1))}{(x + 1) - (x + 1)^3} \]
Simplifying this, we have:
\[ \{x\}^2 = (x + 1)^2 = x^2 + 2x + 1 \]
and
\[ \{x\}^3 = (x + 1)^3 = x^3 + 3x^2 + 3x + 1 \]
The denominator becomes:
\[ (x + 1) - (x^3 + 3x^2 + 3x + 1) = -x^3 - 3x^2 - 2x \]

3. **Calculating L**:
As \( x \to 0^- \):
\[ f(x) \to \frac{\cos^{-1}(1 - (1)^2) \sin^{-1}(1 - 1)}{-0} = \frac{\cos^{-1}(0) \cdot 0}{0} \]
We apply L'Hôpital's Rule to evaluate the limit:
\[ L = \lim_{x \to 0^-} \frac{\cos^{-1}(0) \cdot 0}{-x^3 - 3x^2 - 2x} \]
The numerator approaches \( \frac{\pi}{2} \cdot 0 = 0 \) and the denominator approaches \( 0 \), so we differentiate:
\[ L = \frac{\pi/2}{-2} = \frac{\pi}{4} \]

4. **Finding the right-hand limit (R)**:
For \( x \to 0^+ \) (approaching from the right), we have:
\[ \{x\} = x \]
Thus,
\[ f(x) = \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x - x^3} \]
As \( x \to 0^+ \):
\[ R = \lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x - x^3} \]
Again, applying L'Hôpital's Rule, we differentiate the numerator and denominator:
\[ R = \frac{\pi/2}{1} = \frac{\pi}{\sqrt{2}} \]

5. **Calculating \( \frac{32}{\pi^2}(L^2 + R^2) \)**:
Now substituting \( L \) and \( R \):
\[ L^2 = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16}, \quad R^2 = \left(\frac{\pi}{\sqrt{2}}\right)^2 = \frac{\pi^2}{2} \]
Adding these:
\[ L^2 + R^2 = \frac{\pi^2}{16} + \frac{\pi^2}{2} = \frac{\pi^2}{16} + \frac{8\pi^2}{16} = \frac{9\pi^2}{16} \]
Finally, we calculate:
\[ \frac{32}{\pi^2} \left( \frac{9\pi^2}{16} \right) = \frac{32 \cdot 9}{16} = 18 \]

### Final Answer:
\[ \frac{32}{\pi^2}(L^2 + R^2) = 18 \]