If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, then find the value of a and b.

Let the given polynomial be: p(x) = x³ – 3x² + x + 1

Given, The zeroes of the p(x) are a – b, a, and a + b.

Now, compare the given polynomial equation with the general expression.

px³ + qx² + rx + s = x³ – 3x² + x + 1

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

For the sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

=> - $\frac{\mathrm{q}}{\mathrm{p}}$ = 3a

=>  $\frac{-(- 3)}{1}$= 3a

=> a = 1

So, the zeroes are 1 – b, 1, 1 + b.

For the product of zeroes:

Product of zeroes = 1(1 – b)(1 + b)

=> $\frac{- \mathrm{s}}{\mathrm{p}}$ = 1 – b²

=> $\frac{- 1}{1}$ = 1 – b²

=> b² = 1 + 1 = 2

So, b = √2