Question

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 ...please do mention step by step...and explain the process​..

Answer 1

Step-by-step explanation:

p(x) = ax³ + 3bx² + 3cx + d

Roots are α−β, α, and α+β.

The sum of the roots of a cubic y = ax³ + bx² + cx + d is -b/a.  Therefore:

α−β + α + α+β = -3b/a

3α = -3b/a

α = -b/a

α is a root, so p(α) = 0.

0 = a(α)³ + 3b(α)² + 3c(α) + d

Substitute α = -b/a:

0 = a(-b/a)³ + 3b(-b/a)² + 3c(-b/a) + d

0 = -(b³/a²) + 3b³/a² − 3bc/a + d

0 = -b³ + 3b³ − 3abc + a²d

0 = 2b³ − 3abc + a²d

3abc = 2b³ + a²d