1. **Stating the problem:** Solve the cubic equation $$3x^3 - 26x^2 + 52x - 24 = 0$$ given that the roots are in geometric progression. 2. **Let the roots be:** $$a, ar, ar^2$$ where $$a$$ is the first root and $$r$$ is the common ratio. 3. **Sum of roots:** By Viète's formulas for cubic equations $$ax^3 + bx^2 + cx + d = 0$$ with roots $$ ho_1, ho_2, ho_3$$, $$ ho_1 + ho_2 + ho_3 = -\frac{b}{a}$$ For our equation, $$a=3, b=-26$$, so $$a + ar + ar^2 = \frac{26}{3}$$ Factor out $$a$$: $$a(1 + r + r^2) = \frac{26}{3}$$ 4. **Sum of products of roots taken two at a time:** $$ ho_1\rho_2 + \rho_2\rho_3 + \rho_3\rho_1 = \frac{c}{a}$$ Here, $$c=52$$, so $$a^2(r + r^2 + r^3) = \frac{52}{3}$$ Note that $$r + r^2 + r^3 = r(1 + r + r^2)$$, so $$a^2 r (1 + r + r^2) = \frac{52}{3}$$ 5. **Product of roots:** $$ ho_1 \rho_2 \rho_3 = -\frac{d}{a}$$ Here, $$d = -24$$, so $$a^3 r^3 = \frac{24}{3} = 8$$ 6. **From step 3:** $$a = \frac{26}{3(1 + r + r^2)}$$ 7. **Substitute $$a$$ into step 5:** $$\left(\frac{26}{3(1 + r + r^2)}\right)^3 r^3 = 8$$ Simplify: $$\frac{26^3 r^3}{27 (1 + r + r^2)^3} = 8$$ Multiply both sides by denominator: $$26^3 r^3 = 8 \times 27 (1 + r + r^2)^3$$ Calculate constants: $$26^3 = 17576$$ $$8 \times 27 = 216$$ So: $$17576 r^3 = 216 (1 + r + r^2)^3$$ 8. **Take cube root of both sides:** $$\sqrt[3]{17576} r = \sqrt[3]{216} (1 + r + r^2)$$ Since $$\sqrt[3]{17576} = 26$$ and $$\sqrt[3]{216} = 6$$, $$26 r = 6 (1 + r + r^2)$$ 9. **Rewrite:** $$26 r = 6 + 6 r + 6 r^2$$ Bring all terms to one side: $$6 r^2 + 6 r + 6 - 26 r = 0$$ Simplify: $$6 r^2 - 20 r + 6 = 0$$ Divide entire equation by 2: $$3 r^2 - 10 r + 3 = 0$$ 10. **Solve quadratic for $$r$$:** $$r = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 3 \times 3}}{2 \times 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6}$$ $$r = \frac{10 \pm 8}{6}$$ So, $$r_1 = \frac{18}{6} = 3$$ $$r_2 = \frac{2}{6} = \frac{1}{3}$$ 11. **Find corresponding $$a$$ for each $$r$$ using step 3:** For $$r=3$$: $$a = \frac{26}{3(1 + 3 + 9)} = \frac{26}{3 \times 13} = \frac{26}{39} = \frac{2}{3}$$ For $$r=\frac{1}{3}$$: $$a = \frac{26}{3(1 + \frac{1}{3} + \frac{1}{9})} = \frac{26}{3 \times \frac{13}{9}} = \frac{26}{\frac{39}{9}} = \frac{26 \times 9}{39} = 6$$ 12. **Roots for each case:** - For $$r=3$$: $$a = \frac{2}{3}$$ Roots: $$\frac{2}{3}, 2, 6$$ - For $$r=\frac{1}{3}$$: $$a = 6$$ Roots: $$6, 2, \frac{2}{3}$$ Both sets are the same roots in different order. 13. **Verify roots satisfy the original equation:** The roots are $$\frac{2}{3}, 2, 6$$. 14. **Final answer:** The roots of the equation $$3x^3 - 26x^2 + 52x - 24 = 0$$ with roots in geometric progression are $$\boxed{\frac{2}{3}, 2, 6}$$.