Given that the roots of x^3+3px^2+3qx+r=0 are | vedclass

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(A) Let the roots of the equation $x^3+3px^2+3qx+r=0$ be $\alpha, \beta, \gamma$. Since they are in harmonic progression,their reciprocals $\frac{1}{\

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$2q^3=r(3pq-r)$

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(A) Let the roots of the equation $x^3+3px^2+3qx+r=0$ be $\alpha, \beta, \gamma$. Since they are in harmonic progression,their reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression. Let $\frac{1}{\alpha} = a-d, \frac{1}{\beta} = a, \frac{1}{\gamma} = a+d$. Substituting $x = \frac{1}{y}$ into the original equation,we get $\frac{1}{y^3} + \frac{3p}{y^2} + \frac{3q}{y} + r = 0$,which simplifies to $ry^3 + 3qy^2 + 3py + 1 = 0$. Since the roots of this equation are in arithmetic progression,the sum of the roots is $3a = -\frac{3q}{r}$,so $a = -\frac{q}{r}$. Since $a$ is a root of $ry^3 + 3qy^2 + 3py + 1 = 0$,we have $r(-\frac{q}{r})^3 + 3q(-\frac{q}{r})^2 + 3p(-\frac{q}{r}) + 1 = 0$. $-\frac{q^3}{r^2} + \frac{3q^3}{r^2} - \frac{3pq}{r} + 1 = 0$. Multiplying by $r^2$,we get $-q^3 + 3q^3 - 3pqr + r^2 = 0$,which simplifies to $2q^3 = 3pqr - r^2 = r(3pq-r)$.

Given that the roots of $x^3+3px^2+3qx+r=0$ are in harmonic progression,then:

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