Equating $\mathrm{x}^{2}-\mathrm{x}$ to 0 to find the zeroes, we will get
Since, $x^{3}+x^{2}-a x+b$ is divisible by $x^{2}-x$
Hence, the zeroes of $x^{2}-x$ will satisfy $x^{3}+x^{2}-a x+b$
$\therefore(0)^{3}+0^{2}-\mathrm{a}(0)+\mathrm{b}=0$
$\Rightarrow \mathrm{b}=0$
And
$(1)^{3}+1^{2}-\mathrm{a}(1)+0=0 \quad[\because \mathrm{b}=0]$
$\Rightarrow \mathrm{a}=2$