The condition that the roots of x^3-b x^2+c x | vedclass

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Question

(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$. Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3

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$9 c b=2 b^3+27 d$

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(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$. Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3\alpha$. From the given equation $x^3-b x^2+c x-d=0$,the sum of the roots is $b$. Therefore,$3\alpha = b \Rightarrow \alpha = \frac{b}{3}$. Since $\alpha$ is a root of the equation,it must satisfy $x^3-b x^2+c x-d=0$. Substituting $x = \frac{b}{3}$: $(\frac{b}{3})^3 - b(\frac{b}{3})^2 + c(\frac{b}{3}) - d = 0$ $\frac{b^3}{27} - \frac{b^3}{9} + \frac{bc}{3} - d = 0$ Multiplying the entire equation by $27$: $b^3 - 3b^3 + 9bc - 27d = 0$ $-2b^3 + 9bc - 27d = 0$ $9bc = 2b^3 + 27d$.

The condition that the roots of $x^3-b x^2+c x-d=0$ are in arithmetic progression is

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