The polynomial $x^3-3x^2$ + ax + $b$ is divisible by $(x-1{)}^2$. Then $a+b$ equals:

Divisibility by $(x-1{)}^2$ certainly forces divisibility by $(x-1)$, so $p(1)=1-3+a+b=0$ and $a+b=2$. (Leaving 1 − 3 on the wrong side gives −2.) The full condition pins the polynomial down: $x^3-3x^2+3x-1=(x-1{)}^3$, so $a=3$ and $b=-1$, which indeed add to 2.