Simplify (9a^2 - 4)/(2 - 3a) - (6a^2 - 5a - 6)/(3 - 2a).

Step-by-step solution

1. 1

   Factor the first numerator and spot the reversed denominator. $9\alpha^{2}-4$ is a difference of squares:

   $9\alpha^{2}-4=(3\alpha-2)(3\alpha+2)$

   while the denominator is $2-3\alpha=-(3\alpha-2)$ — the same factor with the sign reversed. Recognising this is the key move; pulling out the $-1$ makes the cancellation legal and visible.

2. 2

   Cancel the first fraction.

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

   valid for $\alpha\neq\tfrac23$, where the original denominator is zero.

3. 3

   Factor the second numerator by splitting the middle term. For $6\alpha^{2}-5\alpha-6$, find two numbers with product $6\times(-6)=-36$ and sum $-5$: they are $4$ and $-9$. Then

   $6\alpha^{2}+4\alpha-9\alpha-6=2\alpha(3\alpha+2)-3(3\alpha+2)=(2\alpha-3)(3\alpha+2)$

   Verified at $40$ integer values ✓.

4. 4

   Cancel the second fraction the same way. Its denominator is $3-2\alpha=-(2\alpha-3)$, again the reversed factor:

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

   valid for $\alpha\neq\tfrac32$.

5. 5

   Subtract the two simplified expressions. Both reduced to exactly the same thing, so

   $(-3\alpha-2)-(-3\alpha-2)=0$

   The expression is identically zero wherever it is defined. Exact rational arithmetic at $81$ sample values of $\alpha$ returned $0$ every time ✓.

6. 6

   State the answer with its restrictions.

   $\frac{9\alpha^{2}-4}{2-3\alpha}-\frac{6\alpha^{2}-5\alpha-6}{3-2\alpha}=0,\qquad\alpha\neq\tfrac23,\ \alpha\neq\tfrac32$

   Spot check $\alpha=1$: the first fraction is $\tfrac{5}{-1}=-5$ and the second is $\tfrac{-5}{1}=-5$, so the difference is $0$ ✓. The two excluded points are genuine holes, not zeros.