It is given that the polynomial f(x)=x³+x²−ax+b is divisible by x²−x which can be rewritten as x(x−1). It means that the given polynomial is divisible by both x and (x−1) that is they both are factors of f(x)=x³+x²−ax+b.

Therefore, x=0 and x=1 are the zeroes of f(x) that is both f(0)=0 and f(1)=0.

Let us first substitute x=0 in f(x)=x³+x²−ax+b as follows:

f(0)=0³+0²−(a×0)+b
⇒0=0³+0²−(a×0)+b
⇒0=0+b
⇒b=0

Now, substitute x=1:

f(1)=1³+1²−(a×1)+b
⇒0=1+1−a+b
⇒0=2−a+b
⇒0=2−a+0 (∵b=0)
⇒a=2

Hence, a=2 and b=0.