Given polynomial x3 – 3x2 + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes
= (a – b) + a + (a + b) = \(\frac{-(-3)}{1}\) = 3
So, 3a = 3 ⇒ a = 1
∴ Sum of the products of its zeroes taken two at a time.
= a(a – b) + a(a + b) + (a + b) (a – b) = \(\frac{1}{1}\) = 1
⇒ a2 – ab + a2 + ab + a2 – b2 = 1
⇒ 3a2 – b2 = 1
So, 3(1)2 – b2 = 1
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = √2 = ± √2
Here, a = 1 and b = ± √2