If the roots of the equation x^3+ax^2+bx+c=0 | vedclass

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(B) Let the roots of the cubic equation be $A-d, A, A+d$. Since the sum of the roots is $-a$,we have $(A-d) + A + (A+d) = -a$,which gives $3A = -a$,so

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$9ab=2a^3+27c$

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(B) Let the roots of the cubic equation be $A-d, A, A+d$. Since the sum of the roots is $-a$,we have $(A-d) + A + (A+d) = -a$,which gives $3A = -a$,so $A = -\frac{a}{3}$. Since $A$ is a root,it must satisfy the equation: $A^3 + aA^2 + bA + c = 0$. Substituting $A = -\frac{a}{3}$: $(-\frac{a}{3})^3 + a(-\frac{a}{3})^2 + b(-\frac{a}{3}) + c = 0$ $-\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + c = 0$ Multiplying by $27$: $-a^3 + 3a^3 - 9ab + 27c = 0$ $2a^3 - 9ab + 27c = 0$ $9ab = 2a^3 + 27c$.

If the roots of the equation $x^3+ax^2+bx+c=0$ are in arithmetic progression,then

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