Question:

Step 1: Factor the given functional equation. We start with the equation: $$f^3(x) - (x^3 + 2)f^2(x) + (2x^3 + 1)f(x) - x^3 = 0$$ Rearranging the terms strategically: $$f^3(x) - x^3 f^2(x) - 2f^2(x) + 2x^3 f(x) + f(x) - x^3 = 0$$ Step 2: Group terms to identify common factors. We group the terms in pairs: $$f^2(x)[f(x) - x^3] - 2f(x)[f(x) - x^3] + 1[f(x) - x^3] = 0$$ Factoring out $(f(x) - x^3)$: $$(f(x) - x^3)[f^2(x) - 2f(x) + 1] = 0$$ Step 3: Simplify the quadratic factor. Recognize that $f^2(x) - 2f(x) + 1$ is a perfect square: $$(f(x) - x^3)(f(x) - 1)^2 = 0$$ This gives us two cases: either $f(x) = x^3$ or $f(x) = 1$. Step 4: Determine which case applies using the invertibility condition. Since $f$ is invertible (one-to-one and onto), the function $f(x) = 1$ for all $x$ cannot be the solution because a constant function is not one-to-one. Therefore: $f(x) = x^3$ Step 5: Calculate $f'(8)$. Taking the derivative of $f(x) = x^3$: $$f'(x) = 3x^2$$ At $x = 8$: $$f'(8) = 3(8)^2 = 3 \cdot 64 = 192$$ Step 6: Find $f^{-1}(8)$ and calculate $(f^{-1})'(8)$. Since $f(x) = x^3$, we have $f^{-1}(x) = \sqrt[3]{x}$. To find $f^{-1}(8)$: we need $f(a) = 8$, which means $a^3 = 8$, so $a = 2$. Therefore: $f^{-1}(8) = 2$ Step 7: Apply the inverse function derivative formula. By the inverse function theorem: $$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$ At $y = 8$: $$(f^{-1})'(8) = \frac{1}{f'(f^{-1}(8))} = \frac{1}{f'(2)} = \frac{1}{3(2)^2} = \frac{1}{3 \cdot 4} = \frac{1}{12}$$ Step 8: Calculate the final product. $$f'(8) \times (f^{-1})'(8) = 192 \times \frac{1}{12} = \frac{192}{12} = 16$$ The value of $f'(8) \times (f^{-1})'(8) = 16$. **Answer: Option 2**