# 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 = $$\pm \sqrt{2}$$