SOLUTION —
As $(a-b), a$ and $(a+b)$ are zeros of $x^3-3 x^2+x+1$,
we have $: a-b+a+a+b=3$
$\Rightarrow 3 a=3$ or, $a=1$ ..... (i)
$a(a-b)+a(a+b)+(a-b)(a+b)=1$
$\Rightarrow 3 a^2-b^2=1$ ........(ii)
and $(a-b) a(a+b)=-1$
$\Rightarrow a\left(a^2-b^2\right)=-1$ ........ (iii)
From (i) and (ii), we have
$b= \pm \sqrt{2}$
Thus, $a=1, b= \pm \sqrt{2}$