Question:

Step 1: Identify the functional equation and recognize it as a cubic in $f(x)$. We are given the equation: $$f^3(x) - (x^3+2)f^2(x) + (2x^3+1)f(x) - x^3 = 0$$ This is a cubic equation in the variable $f(x)$, where $x$ is treated as a parameter. Step 2: Find the roots of the cubic equation by testing candidate values. Let us test $f(x) = 1$: $$1 - (x^3+2)(1) + (2x^3+1)(1) - x^3 = 1 - x^3 - 2 + 2x^3 + 1 - x^3 = 0 \,\checkmark$$ Let us test $f(x) = x^3$: $$x^9 - (x^3+2)x^6 + (2x^3+1)x^3 - x^3 = x^9 - x^9 - 2x^6 + 2x^6 + x^3 - x^3 = 0 \,\checkmark$$ Both $f(x) = 1$ and $f(x) = x^3$ are roots of the cubic. Step 3: Factor the cubic completely. Since we have found two roots, we can write the cubic in factored form. Testing the factorization $(f-1)^2(f-x^3)$: $$(f-1)^2(f-x^3) = (f^2 - 2f + 1)(f - x^3)$$ $$= f^3 - x^3f^2 - 2f^2 + 2x^3f + f - x^3$$ $$= f^3 - (x^3+2)f^2 + (2x^3+1)f - x^3 \,\checkmark$$ Therefore, the cubic factors as: $$[f(x) - 1]^2[f(x) - x^3] = 0$$ Step 4: Determine which root corresponds to an invertible function. The roots are $f(x) = 1$ (with multiplicity 2) and $f(x) = x^3$. Since $f$ must be invertible from $\mathbb{R} \to \mathbb{R}$, the constant function $f(x) = 1$ cannot be invertible (it is not surjective). Therefore: $$f(x) = x^3$$ Step 5: Calculate $f'(8)$. Taking the derivative: $$f'(x) = 3x^2$$ $$f'(8) = 3(8)^2 = 3 \cdot 64 = 192$$ Step 6: Find the inverse function and calculate $(f^{-1})'(8)$. Since $f(x) = x^3$, the inverse function is: $$f^{-1}(x) = x^{1/3}$$ Taking the derivative: $$(f^{-1})'(x) = \frac{1}{3}x^{-2/3}$$ At $x = 8$: $$(f^{-1})'(8) = \frac{1}{3}(8)^{-2/3} = \frac{1}{3} \cdot \frac{1}{(8)^{2/3}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$ Step 7: Calculate the product $f'(8) \times (f^{-1})'(8)$. $$f'(8) \times (f^{-1})'(8) = 192 \times \frac{1}{12} = 16$$ **Final Answer:** The value of $f'(8) \times (f^{-1})'(8) = \boxed{16}$, which corresponds to **Option 2**.