If the zeroes of ax^3 +3bx^2 +3cx +d are in A.P prove that 2b^3 - 3abc

1.	If the zeroes of ax^3 +3bx^2 +3cx +d are in A.P prove that 2b^3 - 3abc+ a^2 d =0
Answer» Given\xa0polynomial is f(x) = ax3 + 3bx2 + 3cx + d.Let p-q, p, and p+q\xa0be the zeros of the polynomial f(x). Then,Sum of the zeros =\xa0{tex}-\\frac{\\text { coefficient of } x^{2}}{\\text { coefficient of } x^{3}}{/tex}p - q\xa0+ p + p + q\xa0=\xa0{tex}\\frac{-3 b}{a}{/tex}{tex}3 p=\\frac{-3 b}{a}{/tex}p =\xa0{tex}\\frac{-b}{a}{/tex}Since p\xa0is a zero of the polynomial f(x).Therefore, f(p) = 0f(p)=ap3 + 3bp2 + 3cp\xa0+ 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 = 0Therefore, 2b3 - 3abc + a2d = 0Hence proved.

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