$\displaystyle a^2(x-1)-9=a(3x-6)$

$\displaystyle x*a(a-3)=a^2-6a+9$

$\displaystyle x*a(a-3)=(a-3)^2$

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

Now... I am sure that solutioin is not defined for a=0, but can I reduce the fraction to:

$\displaystyle x=\frac{(a-3)}{a}$,

without any notes, or do I have to state that reduction of fraction is only posible for $\displaystyle a$ not equal 3? Or in other words, is a=3 also a solution?