1. **State the problem:** Solve the equation $10 = -(x^3 + 3x^2 + 3x) + 3$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to isolate the polynomial. $$10 = -(x^3 + 3x^2 + 3x) + 3$$ Subtract 3 from both sides: $$10 - 3 = -(x^3 + 3x^2 + 3x)$$ $$7 = -(x^3 + 3x^2 + 3x)$$ Multiply both sides by $-1$: $$-7 = x^3 + 3x^2 + 3x$$ 3. **Bring all terms to one side:** $$x^3 + 3x^2 + 3x + 7 = 0$$ 4. **Recognize the cubic form:** The expression $x^3 + 3x^2 + 3x + 1$ is the expansion of $(x+1)^3$. Here, we have $+7$ instead of $+1$, so rewrite as: $$x^3 + 3x^2 + 3x + 1 + 6 = 0$$ $$ (x+1)^3 + 6 = 0$$ 5. **Solve for $(x+1)^3$:** $$(x+1)^3 = -6$$ 6. **Take the cube root:** $$x + 1 = oot 3 ext{ of } -6 = - oot 3 ext{ of } 6$$ 7. **Final solution:** $$x = -1 - oot 3 ext{ of } 6$$ This is the exact real root of the equation. **Answer:** $x = -1 - \sqrt[3]{6}$