Find the sum. Please

Answer:
[tex]\dfrac{2y^2 +12y -8}{y^3-3y+2}[/tex]
Step-by-step explanation:
It usually works to factor the denominators, so you can determine the least common denominator.
[tex]\dfrac{2y}{y^2-2y+1}+\dfrac{8}{y^2+y-2}=\dfrac{2y}{(y-1)^2}+\dfrac{8}{(y-1)(y+2)}\\\\=\dfrac{2y(y+2)}{(y-1)^2(y+2)}+\dfrac{8(y-1)}{(y-1)^2(y+2)}=\dfrac{2y^2+4y+8y-8}{(y-1)^2(y+2)}\\\\=\boxed{\dfrac{2y^2 +12y -8}{y^3-3y+2}}[/tex]