Question

If alpha and beta are the roots of quadratic equation px2 + qx + r = 0, then evaluate alpha cube + beta cube

Answer 1

hello here is your answer by Sujeet yaduvanshi ☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝


GIVEN THAT:)
Zeroes of polynomial px²+qx+r=0

alpha and beta are zeroes of polynomial


then,

a+b=-b/a
a+b=-q/p


Again..

ab=c/a
ab=r/p

then,

value of

a³+b³

(a+b)(a²+b²-ab)

(-q/p){(a+b)²-2ab -ab

(-q/p){(-q/p)²-2*r/p -r/p
(-q/p) [(q²/p²-2r/p -r/p

Next step see in attachment


that's all

Answer 2

Answer:

Answer is

(3pr-q^3)/(p^3)