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

Step-by-step explanation:

(x - (a - b))·(x - a)·(x - (a + b))

= (x - a + b)·(x - a)·(x - a - b)

= ((x - a) + b)·(x - a)·((x - a) - b)

= ((x - a)^2 - b^2)·(x - a)

= (x^2 - 2·a·x + a^2 - b^2)·(x - a)

= x^3 - 2·a·x^2 + a^2·x - b^2·x - x^2·a + 2·a^2·x - a^3 + a·b^2

= x^3 - 3·a·x^2 + 3·a^2·x - b^2·x - a^3 + a·b^2

= x^3 - 3·a·x^2 + (3·a^2 - b^2)·x + (a·b^2 - a^3)

= x^3 - 3·x^2 + 1·x + 1

So

- 3·a = - 3 --> a = 1

3·a^2 - b^2 = 1

3 - b^2 = 1

3 - 1 = b^2

2 = b^2

b = ± √2

3; 9 is 8 to -4

step-by-step explanation: