[tex]\dfrac{3a^2-6a+3}{a^2-1}\\\\=\dfrac{3(a^2 -2a+1)}{(a+1)(a-1)}\\\\=\dfrac{3(a-1)^2}{(a+1)(a-1)}\\\\=\dfrac{3(a-1)}{a+1}\\\\=\dfrac{3a-3}{a+1} ~~~\text{for}~ a\neq 1[/tex]
