Question

If alpha beta gamma are the zeros of polynomial P(x) =x^3-6x^2+11x-6 find the value of alpha^3+ beta^3 + gamma^3

Answer 1

factorize x^3-6x^2+11x-6 = (x - 1)(x - 2)(x - 3)

Steps for Factorization:

i) Factors of 6 - 1, 2, 3, -1, -2, -3

ii) Divide x^3-6x^2+11x-6 by (x - 1) ---> remainder = 0; Quotient = x^2 -5x + 6

iii) Factor  x^2 -5x + 6 by grouping

    x^2 - 5x + 6 = x^2 -2x - 3x + 6 = x (x - 2) - 3 (x - 2) = (x - 3)(x - 2)

So the zeroes of the equation are....

x - 1 = 0 => x = 1 (alpha)

x -2 = 0 => x = 2 ( beta)

x - 3 = 0 => x = 3 (gamma)

alpha^3 + beta^3 + gamma^3

=> 1^3 + 2^3 + 3^3

=> 1 + 8 + 27

=> 36

The answer is 36.

Hope this helps ;)