If the zeroes of the polynomial \(x^3 – 3x^2 + x + 1\) are a – b, a and a + b, find a and b.

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 = \(\pm \sqrt{2}\)