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

We know that,
Sum of zeroes = -(Cofficient of x²)/Cofficient of x³)
Therefore,
(a – b) + a + (a + b) = (-(-3))/1
⇒ 3a = 3
⇒ a = 1
Product of zeroes = -(Constant term)/(Cofficient of x³)
Therefore,
(a – b)(a)(a + b) = -(1)/1
⇒ (1 – b)1(1 + b) = -1 [Because a = 1]
⇒ 1 – b² = – 1
⇒ b² = 2
⇒ b = +- √2
Hence, a = 1 and b = +- √2

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