using the relationship between the zeroes of he quadratic polynomial.

Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$

$\therefore a-b+a+a+b=\frac{-(-3)}{1}$

$\Rightarrow 3 \mathrm{a}=3$ $\Rightarrow \mathrm{a}=1$

Now, Product of zeroes $=\frac{-(\text { constant term })}{\text { coefficient of } x^{3}}$

$\therefore(a-b)(a)(a+b)=\frac{-1}{1}$

$\Rightarrow(1-\mathrm{b})(1)(1+\mathrm{b})=-1 \quad[\because \mathrm{a}=1]$

$\Rightarrow 1-\mathrm{b}^{2}=-1$

$\Rightarrow b^{2}=2$

$\Rightarrow \mathrm{b}=\pm \sqrt{2}$