We need to solve the equation $\frac{x-1}{a-3} - \frac{ax}{a^2+3a+9} = \frac{21x-a^2}{a^3-27}$ for $x$.

1. Problem Description

We need to solve the equation

$\frac{x-1}{a-3} - \frac{ax}{a^2+3a+9} = \frac{21x-a^2}{a^3-27}$ for $x$.

2. Solution Steps

First, we factor $a^3 - 27$. We have

$a^3 - 27 = a^3 - 3^3 = (a-3)(a^2+3a+9)$.

So the equation becomes

$\frac{x-1}{a-3} - \frac{ax}{a^2+3a+9} = \frac{21x-a^2}{(a-3)(a^2+3a+9)}$.

We multiply both sides by $(a-3)(a^2+3a+9)$ to get rid of the denominators.

$(x-1)(a^2+3a+9) - ax(a-3) = 21x-a^2$.

Expanding the terms:

$x(a^2+3a+9) - (a^2+3a+9) - a^2x + 3ax = 21x - a^2$.

$xa^2+3ax+9x - a^2 - 3a - 9 - a^2x + 3ax = 21x - a^2$.

Now, we simplify by cancelling $xa^2$ and $-a^2x$, and collect the terms containing $x$ on one side:

$6ax+9x - a^2 - 3a - 9 = 21x - a^2$.

$6ax + 9x - 21x = -a^2 + a^2 + 3a + 9$.

$6ax - 12x = 3a + 9$.

Now, we factor out $x$ from the left side:

$x(6a - 12) = 3a + 9$.

$x = \frac{3a+9}{6a-12}$.

$x = \frac{3(a+3)}{6(a-2)}$.

$x = \frac{a+3}{2(a-2)}$.

3. Final Answer

$x = \frac{a+3}{2(a-2)}$