If the roots of the cubic equation ax^3 + bx^ | vedclass

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Question

(A) Let the roots of the equation $ax^3 + bx^2 + cx + d = 0$ be $\frac{A}{R}, A, AR$.The product of the roots is given by $\frac{-d}{a}$.So,$(\frac{A}

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$c^3a = b^3d$

Vedclass · Answer

(A) Let the roots of the equation $ax^3 + bx^2 + cx + d = 0$ be $\frac{A}{R}, A, AR$.The product of the roots is given by $\frac{-d}{a}$.So,$(\frac{A}{R}) \cdot A \cdot (AR) = A^3 = -\frac{d}{a}$.Since $A$ is a root of the equation,it must satisfy $aA^3 + bA^2 + cA + d = 0$.Substituting $A^3 = -\frac{d}{a}$,we get $a(-\frac{d}{a}) + bA^2 + cA + d = 0$,which simplifies to $bA^2 + cA = 0$.Since $A 
eq 0$ (assuming $d 
eq 0$),we have $bA + c = 0$,so $A = -\frac{c}{b}$.Substituting $A = -\frac{c}{b}$ into $A^3 = -\frac{d}{a}$,we get $(-\frac{c}{b})^3 = -\frac{d}{a}$.This implies $-\frac{c^3}{b^3} = -\frac{d}{a}$,which simplifies to $c^3a = b^3d$.

If the roots of the cubic equation $ax^3 + bx^2 + cx + d = 0$ are in $G.P.$,then

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