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If the zeroes of the polynomial $$x^3 - 3x^2 + x + 1$$ are $$a-b , a , a+b$$ find $$a$$ and $$b$$.

Let the three zeroes be p,q and r

We know:

$$p + q + r$$ = $${{-b}\over{a}} =$$ ............(i)
$$pq + qr + rp$$ = $${{c}\over{a}}$$..........(ii)
$$pqr$$ = $${{-d}\over{a}}$$.................(iii)

On checking equation (i):=> $$a - b + a + a + b = 3a$$

=>$$3a = {{-b}\over{a}} = {{-(-3)}\over{1}}$$
=> $$3a = 3$$
=> $$a = 1$$

On checking equation (ii):=> $$(a - b) × a + a × (a + b) + (a + b) × (a - b)$$

=>$$a^2 - ab + a^2 + ab + a^2 - b^2 = {{c}\over{a}} = {{1}\over{1}} = 1$$
=> $$3a^2 - b^2 = 1$$
=> $$3(1)^2 - b^2 = 1 (a = 1)$$
=>$$3 - b^2 = 1$$
=> $$b^2 = 4$$
=> $$b = +2, -2$$

Therefore, $$a = 1, b = +2, -2$$