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(C) Given that $a, b, c$ are in $H.P.$Therefore,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$This implies $\frac{1}{b} - \fr

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$\frac{3}{b^2} - \frac{2}{ab}$

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(C) Given that $a, b, c$ are in $H.P.$Therefore,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.Let the expression be $E = \left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$.Using the relation $\frac{1}{c} - \frac{1}{a} = \frac{1}{b} - \frac{2}{a}$ (from $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$),the first bracket becomes $\left( \frac{1}{b} + (\frac{2}{b} - \frac{1}{a}) - \frac{1}{a} \right) = \left( \frac{3}{b} - \frac{2}{a} \right)$.Using the relation $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$,the second bracket becomes $\left( \frac{2}{b} - \frac{1}{b} \right) = \frac{1}{b}$.Thus,$E = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

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