1. **State the problem:** Solve the cubic equation $$2a^3 - 13a^2 + 27a - 18 = 0.$$\n\n2. **Formula and approach:** To solve a cubic polynomial equation, we try to find rational roots using the Rational Root Theorem, then factor the polynomial and solve the resulting quadratic if needed.\n\n3. **Check possible rational roots:** Possible roots are factors of the constant term 18 divided by factors of the leading coefficient 2, i.e., $$\pm1, \pm2, \pm3, \pm6, \pm9, \pm18, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{9}{2}.$$\n\n4. **Test root $a=1$: Substitute into the polynomial:**\n$$2(1)^3 - 13(1)^2 + 27(1) - 18 = 2 - 13 + 27 - 18 = -11 + 27 - 18 = 16 - 18 = -2 \neq 0.$$\n\n5. **Test root $a=3$: Substitute:**\n$$2(3)^3 - 13(3)^2 + 27(3) - 18 = 2(27) - 13(9) + 81 - 18 = 54 - 117 + 81 - 18 = (54 - 117) + (81 - 18) = -63 + 63 = 0.$$\nSo, $a=3$ is a root.\n\n6. **Divide the polynomial by $(a - 3)$ using polynomial division or synthetic division:**\n\nPolynomial: $$2a^3 - 13a^2 + 27a - 18$$\nDivide by $(a - 3)$:\n\nUsing synthetic division with root 3:\n\nCoefficients: 2 | -13 | 27 | -18\n\nBring down 2: 2\nMultiply 2*3=6, add to -13: -7\nMultiply -7*3=-21, add to 27: 6\nMultiply 6*3=18, add to -18: 0\n\nQuotient polynomial: $$2a^2 - 7a + 6$$\n\n7. **Solve quadratic equation:** $$2a^2 - 7a + 6 = 0.$$\n\nUse quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 2 \times 6}}{2 \times 2} = \frac{7 \pm \sqrt{49 - 48}}{4} = \frac{7 \pm 1}{4}.$$\n\n8. **Calculate roots:**\n$$a_1 = \frac{7 + 1}{4} = \frac{8}{4} = 2,$$\n$$a_2 = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2}.$$\n\n9. **Final roots:** $$a = 3, 2, \frac{3}{2}.$$