If the zeroes of the cubic polynomial x^{3} - 6x^{2} + 10 are of the form a, a + b and a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Let P(x) = x^{3} - 6x^{2} + 10

And (a), (a + b) and (a + 2b) are the zeroes of P(x).

Sum of the zeroes = - (coefficient of x^{2}) ÷ coefficient of x^{3}

α + β + γ = - b/a

a + (a + b) + (a + 2b) = - (- 6)

= 3a + 3b = 6

= a + b = 2

⇒ a = 2 - b (1)

Product of all the zeroes = - (constant term) ÷ coefficient of x^{3}

αβγ = - d/a

a (a + b)(a + 2b) = - 10

= (2 - b) (2) (2 + b) = - 10

= (2 - b) (2 + b) = - 5

= 4 - b^{2} = - 5

⇒ b^{2} = 9

⇒ b = 3

When b = 3, a = 2 - 3 = - 1 (from (1))

⇒ a = - 1

When b = - 3,a = 2 - (- 3) = 5 (from(1))

⇒ a = 5

Case1: when a = - 1 and b = 3

The zeroes of the polynomial are:

a = - 1 a + b = - 1 + 3 = 2 a + 2b = - 1 + 2(3) = 5

⇒ - 1, 2 and 5 are the zeroes

Case2: when a = 5,b = - 3

The zeroes of the polynomial are:

a = 5 a + b = 5 - 3 = 2 a + 2b = 5 - 2(3) = - 1

⇒ - 1, 2 and 5 are the zeroes

By both the cases the zeroes of the polynomial is - 1, 2, 5

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