We are given with the polynomial here,

p(x) = x³-3x²+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px³+qx²+rx+s = x³-3x²+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = √2

Hence,1-√2, 1 ,1+√2 are the zeroes of x³-3x²+x+1.