If the zeroes of the polynomial

The given polynomial $=x^{3}-3 x^{2}+x+1$ and its roots are $(a-b), a$ and $(a+b)$.

Comparing the given polynomial with $A x^{3}+B x^{2}+C x+D$, we have :

$A=1, B=-3, C=1$ and $D=1$

Now, $(a-b)+a+(a+b)=\frac{-B}{A}$

$=>3 a=-\frac{-3}{1}$

$=>a=1$

Also, $(a-b) \times a \times(a+b)=\frac{-D}{A}$

$=>a\left(a^{2}-b^{2}\right)=\frac{-1}{1}$

$=>1\left(1^{2}-b^{2}\right)=-1$

$=>1-b^{2}=-1$

$=>b^{2}=2$

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

$\therefore a=1$ and $b=\pm \sqrt{2}$