1. **State the problem:** We are given the cubic equation $$y=x^3 - 4x^2 + x + 8$$ and asked to find the values of $x$ when $$y=2$$. 2. **Set the equation equal to 2:** $$x^3 - 4x^2 + x + 8 = 2$$ 3. **Subtract 2 from both sides to set equation to 0:** $$x^3 - 4x^2 + x + 8 - 2 = 0$$ $$x^3 - 4x^2 + x + 6 = 0$$ 4. **Try to find rational roots using trial (possible roots are factors of 6):** Test $x=1$: $$1 - 4 + 1 + 6 = 4 \neq 0$$ Test $x=-1$: $$-1 - 4 + (-1) + 6 = 0$$ So $x=-1$ is a root. 5. **Divide the polynomial by $(x+1)$ using polynomial division or synthetic division:** $$\frac{x^3 - 4x^2 + x + 6}{x+1} = x^2 - 5x + 6$$ 6. **Factor the quadratic:** $$x^2 - 5x + 6 = (x-2)(x-3)$$ 7. **Write all solutions:** The solutions to $$x^3 - 4x^2 + x + 6 = 0$$ are $$x = -1, 2, 3$$. **Final Answer:** When $$y=2$$, $$x=-1, 2, 3$$.