Factorise completely $2x^3 - 3x^2 - 5x + 6$. Hence, solve $2e^{3y} - 3e^{2y} - 5e^y + 6 = 0$.

1. Use the Factor Theorem to find a linear factor of the polynomial.

Let $P(x) = 2x^3 - 3x^2 - 5x + 6$.
By trial and error,
$P(1) = 2(1)^3 - 3(1)^2 - 5(1) + 6 = 2 - 3 - 5 + 6 = 0$.
Since $P(1) = 0$, $(x - 1)$ is a factor.

2. Find the quadratic factor by comparing coefficients or polynomial division.

Let $2x^3 - 3x^2 - 5x + 6 = (x - 1)(2x^2 + bx - 6)$.
Comparing the coefficients of $x^2$:
$-2 + b = -3 \implies b = -1$.
So the quadratic factor is $2x^2 - x - 6$.

3. Factorise the quadratic factor completely.

$2x^2 - x - 6 = (2x + 3)(x - 2)$
Therefore, $2x^3 - 3x^2 - 5x + 6 = (x - 1)(x - 2)(2x + 3)$.

1. Use the factorised form from part (i) by substituting $x = e^y$

From (i), $2x^3 - 3x^2 - 5x + 6 = (x-1)(x-2)(2x+3)$.
Let $x = e^y$.
The equation becomes $(e^y - 1)(e^y - 2)(2e^y + 3) = 0$.

2. Solve for $e^y$ and reject invalid solutions

$e^y = 1$, $e^y = 2$, or $e^y = -\frac{3}{2}$.
Since $e^y > 0$ for all real $y$, $e^y = -\frac{3}{2}$ has no real solution.

3. Solve for $y$

For $e^y = 1$, $y = \ln 1 = 0$.
For $e^y = 2$, $y = \ln 2$.