For constants A, B, C we have:
(3x²+16x+15)/(x+3)³≡A/(x+3)³+B/(x+3)²+C/(x+3).
Therefore, we have:
3x²+16x+15≡A+B(x+3)+C(x²+6x+9)=A+Bx+3B+Cx²+6Cx+9C.
Equating terms:
x² term: C=3
x term: B+6C=16, so B=16-6C=-2
constant: A+3B+9C=15, A=15+6-27=-6.
So partial fractions are:
3/(x+3)-2/(x+3)²-6/(x+3)³