1. **State the problem:** Find the roots of the cubic equation $$-4x^{3}+12x^{2}-6x+1=0$$. 2. **Rewrite the equation:** Multiply both sides by $$-1$$ to simplify the leading coefficient: $$\cancel{-4}x^{3}+\cancel{-12}x^{2}+\cancel{6}x-\cancel{1}=0 \implies 4x^{3}-12x^{2}+6x-1=0$$ 3. **Use the Rational Root Theorem:** Possible rational roots are factors of the constant term divided by factors of the leading coefficient, i.e., $$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$$. 4. **Test possible roots:** Evaluate the polynomial at these values. - At $$x=\frac{1}{2}$$: $$4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 1 = 4\cdot\frac{1}{8} - 12\cdot\frac{1}{4} + 3 - 1 = \frac{1}{2} - 3 + 3 - 1 = -\frac{1}{2} \neq 0$$ - At $$x=1$$: $$4(1)^3 - 12(1)^2 + 6(1) - 1 = 4 - 12 + 6 - 1 = -3 \neq 0$$ - At $$x=\frac{1}{4}$$: $$4\left(\frac{1}{4}\right)^3 - 12\left(\frac{1}{4}\right)^2 + 6\left(\frac{1}{4}\right) - 1 = 4\cdot\frac{1}{64} - 12\cdot\frac{1}{16} + \frac{3}{2} - 1 = \frac{1}{16} - \frac{3}{4} + \frac{3}{2} - 1 = 0$$ 5. **Factor out $$\left(x - \frac{1}{4}\right)$$:** Use polynomial division or synthetic division to divide the cubic by $$\left(x - \frac{1}{4}\right)$$. 6. **Perform synthetic division:** Coefficients: $$4, -12, 6, -1$$ Divide by root $$x=\frac{1}{4}$$: - Bring down 4 - Multiply 4 by $$\frac{1}{4}$$ = 1, add to -12 = -11 - Multiply -11 by $$\frac{1}{4}$$ = -\frac{11}{4}, add to 6 = $$6 - \frac{11}{4} = \frac{24}{4} - \frac{11}{4} = \frac{13}{4}$$ - Multiply $$\frac{13}{4}$$ by $$\frac{1}{4}$$ = $$\frac{13}{16}$$, add to -1 = $$-1 + \frac{13}{16} = -\frac{16}{16} + \frac{13}{16} = -\frac{3}{16}$$ (not zero, so re-check step 4) 7. **Re-evaluate step 4 for $$x=\frac{1}{4}$$:** $$4\left(\frac{1}{4}\right)^3 - 12\left(\frac{1}{4}\right)^2 + 6\left(\frac{1}{4}\right) - 1 = 4\cdot\frac{1}{64} - 12\cdot\frac{1}{16} + \frac{3}{2} - 1 = \frac{1}{16} - \frac{3}{4} + \frac{3}{2} - 1$$ Calculate stepwise: $$\frac{1}{16} - \frac{3}{4} = \frac{1}{16} - \frac{12}{16} = -\frac{11}{16}$$ $$-\frac{11}{16} + \frac{3}{2} = -\frac{11}{16} + \frac{24}{16} = \frac{13}{16}$$ $$\frac{13}{16} - 1 = \frac{13}{16} - \frac{16}{16} = -\frac{3}{16} \neq 0$$ So $$x=\frac{1}{4}$$ is not a root. 8. **Try $$x=\frac{1}{2}$$ again:** $$4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 1 = 4\cdot\frac{1}{8} - 12\cdot\frac{1}{4} + 3 - 1 = \frac{1}{2} - 3 + 3 - 1 = -\frac{1}{2} \neq 0$$ 9. **Try $$x=1$$:** $$4 - 12 + 6 - 1 = -3 \neq 0$$ 10. **Try $$x=\frac{3}{2}$$:** $$4\left(\frac{3}{2}\right)^3 - 12\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) - 1 = 4\cdot\frac{27}{8} - 12\cdot\frac{9}{4} + 9 - 1 = 13.5 - 27 + 9 - 1 = -5.5 \neq 0$$ 11. **Use depressed cubic substitution:** Let $$x = y + 1$$ to simplify the cubic. 12. **Rewrite equation in terms of $$y$$:** $$4(y+1)^3 - 12(y+1)^2 + 6(y+1) - 1 = 0$$ Expand: $$4(y^3 + 3y^2 + 3y + 1) - 12(y^2 + 2y + 1) + 6y + 6 - 1 = 0$$ $$4y^3 + 12y^2 + 12y + 4 - 12y^2 - 24y - 12 + 6y + 6 - 1 = 0$$ Simplify: $$4y^3 + (12y^2 - 12y^2) + (12y - 24y + 6y) + (4 - 12 + 6 - 1) = 0$$ $$4y^3 - 6y - 3 = 0$$ 13. **Divide entire equation by 4:** $$y^3 - \frac{3}{2}y - \frac{3}{4} = 0$$ 14. **Use Cardano's formula for depressed cubic:** General form: $$y^3 + py + q = 0$$ with $$p = -\frac{3}{2}$$ and $$q = -\frac{3}{4}$$. 15. **Calculate discriminant:** $$\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 = \left(-\frac{3}{8}\right)^2 + \left(-\frac{1}{2}\right)^3 = \frac{9}{64} - \frac{1}{8} = \frac{9}{64} - \frac{8}{64} = \frac{1}{64} > 0$$ 16. **Since $$\Delta > 0$$, one real root exists:** $$y = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}}$$ 17. **Calculate:** $$-\frac{q}{2} = \frac{3}{8}$$ $$\sqrt{\Delta} = \frac{1}{8}$$ 18. **Compute cube roots:** $$\sqrt[3]{\frac{3}{8} + \frac{1}{8}} = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}$$ $$\sqrt[3]{\frac{3}{8} - \frac{1}{8}} = \sqrt[3]{\frac{1}{4}} = \frac{1}{\sqrt[3]{4}}$$ 19. **Sum roots:** $$y = \frac{1}{\sqrt[3]{2}} + \frac{1}{\sqrt[3]{4}}$$ 20. **Convert back to $$x$$:** $$x = y + 1 = 1 + \frac{1}{\sqrt[3]{2}} + \frac{1}{\sqrt[3]{4}}$$ **Final answer:** $$\boxed{x = 1 + \frac{1}{\sqrt[3]{2}} + \frac{1}{\sqrt[3]{4}}}$$