If the roots of the equation x^{3}+a x^{2}+b | vedclass

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(D) Given equation is $x^{3}+a x^{2}+b x+c=0$.Let the roots be $(\alpha, \beta, \gamma)$. Since they are in $AP$,we have $2 \beta = \alpha + \gamma$.F

Vedclass · Accepted Answer

$-27$

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(D) Given equation is $x^{3}+a x^{2}+b x+c=0$.Let the roots be $(\alpha, \beta, \gamma)$. Since they are in $AP$,we have $2 \beta = \alpha + \gamma$.From the sum of roots,$\alpha + \beta + \gamma = -a$.Substituting $\alpha + \gamma = 2 \beta$,we get $3 \beta = -a$,so $\beta = -\frac{a}{3}$.Since $\beta$ is a root,it satisfies the equation:$(-\frac{a}{3})^{3} + a(-\frac{a}{3})^{2} + b(-\frac{a}{3}) + c = 0$.$-\frac{a^{3}}{27} + \frac{a^{3}}{9} - \frac{ab}{3} + c = 0$.Multiplying by $27$,we get $-a^{3} + 3a^{3} - 9ab + 27c = 0$.$2a^{3} - 9ab + 27c = 0$.Therefore,$2a^{3} - 9ab = -27c$.

If the roots of the equation $x^{3}+a x^{2}+b x+c=0$ are in $AP$,then $2 a^{3}-9 a b$ is equal to (in $c$)

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