If the zeros of the polynomial

By using the relationship between the zeroes of the cubic ploynomial.

We have, Sum of zeroes $=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficent of } x^{3}}$

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

$\Rightarrow 3 a=3$

$\Rightarrow a=1$

Now, Product of zeros $=\frac{-(\text { constant } t e r m)}{\text { coefficent of } x^{3}}$

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

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

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

$\Rightarrow b^{2}=2$

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