If the zeros of the polynomial $x^3-3x^2+x+1$ are $a-b,a,a+b$, then the value of $a+b$ is

Comparing the given polynomial $x^3-3x^2+x+1$  with $A{x}^3+B{x}^2+Cx+D$. we get
$A=1,B=-3,C=1,D=1$
Here, the zeros of the given polynomial  are $a-b,a,a+b$.
Sum of zeros = $-\frac{B}{A}=-\frac{-3}{1}=3$
$\therefore a-b+a+a+b=3$.
$\Rightarrow 3a=3$.
$\Rightarrow a=1$.
Product of zeros = $-\frac{D}{A}=-\frac{1}{1}=-1$
$\therefore (a-b)(a)(a+b)=-1$.
$\Rightarrow a(a^2-b^2)=-1$.
since, $a=1$
$\Rightarrow (1^2-b^2)=-1$.
$\Rightarrow b^2=2$
$\Rightarrow b=\sqrt{2}$ or $-\sqrt{2}$
value of $a+b=\sqrt{2}+1$ or $1-\sqrt{2}$