If the zeroes of the polynomial $x^{3}-3x^{2}+x+1$ are $a-b, a, a+b$,find $a$ and $b$.

(A) Given polynomial is $p(x) = x^{3}-3x^{2}+x+1$.
The zeroes are given as $a-b, a, a+b$.
Comparing the polynomial with the standard form $Ax^{3}+Bx^{2}+Cx+D$,we get $A=1, B=-3, C=1, D=1$.
Using the relationship between zeroes and coefficients:
Sum of zeroes $= (a-b) + a + (a+b) = -B/A$.
$3a = -(-3)/1 = 3$.
$a = 1$.
Product of zeroes $= (a-b)(a)(a+b) = -D/A$.
$(1-b)(1)(1+b) = -1/1$.
$1-b^{2} = -1$.
$b^{2} = 2$.
$b = \pm\sqrt{2}$.
Thus,$a=1$ and $b=\pm\sqrt{2}$.