if the zeroes of the polynomial, p(x)= ax^3 + 3bx^2 + 3cx + d = 0 be alpha - beta, alpha and alpha + beta. Prove that 3abc = a^2d + 2b^3 ...I hope you help..

Step-by-step explanation:

According to question

α−d+α+α+d=−3ba

3α=−3ba

α=−ba

α2−dd+α2+αd+α2−d2=3a

3α2−d2=3ca

3b2a2−h2=3ca

d2=3b2a2−3ca=3b2−3aca2

α(α2−d2)=−da

−ba(b2a2−(3b2−3aca2))=−da

b(b2−3b2+3aca2)=d

−2b3+3ab=a2d

2b3−3abc+a2d=0

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step-by-step explanation: