Solution :
Since (a – b), a and (a + b) are the zeroes of the polynomials \(x^3 – 3x^2 + x + 1\), therefore
(a – b) + a + (a + b) = \(-(-3)\over 1\) = 3
So, 3a = 3 \(\implies\) a = 1
(a – b)a + a(a + b) + (a + b)(a – b) = \(1\over 1\)
\(\implies\) \(a^2 – ab + a^2 + ab + a^2 – b^2\) = 1
\(\implies\) \(3a^2 – b^2\) = 1
So, \(3(1)^2 – b^2\) = 1
\(\implies\) \(b^2\) = 2 or b = \(\pm \sqrt{2}\)
Hence, a = 1 and b = \(\pm \sqrt{2}\)
