1. **Stating the problem:** Solve the cubic equation $$3x^2 - 6 - x^3 = 0$$ for $x$. 2. **Rewrite the equation:** Rearrange terms to standard polynomial form: $$-x^3 + 3x^2 - 6 = 0$$ Multiply both sides by $-1$ to simplify: $$x^3 - 3x^2 + 6 = 0$$ 3. **Try to find rational roots using the Rational Root Theorem:** Possible roots are factors of 6 (constant term) over factors of 1 (leading coefficient), i.e., $\pm1, \pm2, \pm3, \pm6$. 4. **Test $x=1$:** $$1^3 - 3(1)^2 + 6 = 1 - 3 + 6 = 4 \neq 0$$ 5. **Test $x=2$:** $$2^3 - 3(2)^2 + 6 = 8 - 12 + 6 = 2 \neq 0$$ 6. **Test $x=3$:** $$3^3 - 3(3)^2 + 6 = 27 - 27 + 6 = 6 \neq 0$$ 7. **Test $x=-1$:** $$(-1)^3 - 3(-1)^2 + 6 = -1 - 3 + 6 = 2 \neq 0$$ 8. **Test $x=-2$:** $$(-2)^3 - 3(-2)^2 + 6 = -8 - 12 + 6 = -14 \neq 0$$ 9. **Test $x=-3$:** $$(-3)^3 - 3(-3)^2 + 6 = -27 - 27 + 6 = -48 \neq 0$$ 10. **Test $x=-6$:** $$(-6)^3 - 3(-6)^2 + 6 = -216 - 108 + 6 = -318 \neq 0$$ 11. Since no rational roots are found, use the depressed cubic substitution or numerical methods. 12. **Rewrite original equation:** $$-x^3 + 3x^2 - 6 = 0$$ Multiply both sides by $-1$: $$x^3 - 3x^2 + 6 = 0$$ 13. **Use substitution $x = y + \frac{b}{3a}$ to remove quadratic term:** Here, $a=1$, $b=-3$, so $$x = y + \frac{3}{3} = y + 1$$ 14. Substitute into equation: $$ (y+1)^3 - 3(y+1)^2 + 6 = 0$$ 15. Expand: $$ y^3 + 3y^2 + 3y + 1 - 3(y^2 + 2y + 1) + 6 = 0$$ 16. Simplify: $$ y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 + 6 = 0$$ $$ y^3 + (3y^2 - 3y^2) + (3y - 6y) + (1 - 3 + 6) = 0$$ $$ y^3 - 3y + 4 = 0$$ 17. **Solve depressed cubic:** $$ y^3 - 3y + 4 = 0$$ 18. Use the cubic formula or try to find roots by inspection. 19. Test $y=1$: $$1 - 3 + 4 = 2 \neq 0$$ 20. Test $y=-1$: $$-1 + 3 + 4 = 6 \neq 0$$ 21. Test $y=2$: $$8 - 6 + 4 = 6 \neq 0$$ 22. Test $y=-2$: $$-8 + 6 + 4 = 2 \neq 0$$ 23. Since no simple roots, use the cubic formula: For equation $$y^3 + py + q = 0$$ with $p = -3$, $q = 4$. Discriminant: $$\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 = \left(\frac{4}{2}\right)^2 + \left(\frac{-3}{3}\right)^3 = 2^2 + (-1)^3 = 4 - 1 = 3 > 0$$ 24. Since $\Delta > 0$, one real root and two complex roots. 25. Real root: $$ y = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} = \sqrt[3]{-2 + \sqrt{3}} + \sqrt[3]{-2 - \sqrt{3}}$$ 26. Approximate numerically: $$ \sqrt{3} \approx 1.732$$ $$ y \approx \sqrt[3]{-0.268} + \sqrt[3]{-3.732} \approx -0.645 - 1.557 = -2.202$$ 27. Recall substitution $x = y + 1$: $$ x \approx -2.202 + 1 = -1.202$$ 28. **Final answer:** The equation has one real root approximately $$ x \approx -1.202$$ and two complex roots (not calculated here).