If the zeroes of ax^3 +3bx^2 …

Given polynomial is f(x) = ax³ + 3bx² + 3cx + d.
Let p-q, p, and p+q be the zeros of the polynomial f(x). Then,
Sum of the zeros = {tex}-\frac{\text { coefficient of } x^{2}}{\text { coefficient of } x^{3}}{/tex}
p - q + p + p + q = {tex}\frac{-3 b}{a}{/tex}
{tex}3 p=\frac{-3 b}{a}{/tex}
p = {tex}\frac{-b}{a}{/tex}
Since p is a zero of the polynomial f(x).
Therefore, f(p) = 0
f(p)=ap³ + 3bp² + 3cp + d = 0
{tex}\Rightarrow{/tex}{tex}a\left(\frac{-b}{a}\right)^{3}+3 b \times\left(\frac{-b}{a}\right)^{2}+3 c \times\left(\frac{-b}{a}\right)+d=0{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{-b^{3}}{a^{2}}+\frac{3 b^{3}}{a^{2}}-\frac{3 c b}{a}+d=0{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{-b^{3}+3 b^{3}-3 a b c+a^{2} d}{a^{2}}=0{/tex}
{tex}\Rightarrow{/tex}2b³ - 3abc +a²d = 0
Therefore, 2b³ - 3abc + a²d = 0
Hence proved.