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(A) Since $a, b, c$ are in $H.P.$,their reciprocals $1/a, 1/b, 1/c$ are in $A.P.$This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,

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$\frac{2}{{bc}} - \frac{1}{{{b^2}}}$

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(A) Since $a, b, c$ are in $H.P.$,their reciprocals $1/a, 1/b, 1/c$ are in $A.P.$This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$     .....$(1)$Now,consider the expression:$E = \left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right)\left( {\frac{1}{c} + \frac{1}{a} - \frac{1}{b}} \right)$From equation $(1)$,we have $\frac{1}{c} - \frac{1}{a} = \frac{2}{b} - \frac{2}{a}$ (not directly useful) or $\frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$.Let's substitute $\frac{1}{a} = \frac{2}{b} - \frac{1}{c}$ into the first bracket:$\left( \frac{1}{b} + \frac{1}{c} - (\frac{2}{b} - \frac{1}{c}) \right) = \left( \frac{2}{c} - \frac{1}{b} \right)$Substitute $\frac{1}{a} = \frac{2}{b} - \frac{1}{c}$ into the second bracket:$\left( \frac{1}{c} + (\frac{2}{b} - \frac{1}{c}) - \frac{1}{b} \right) = \left( \frac{1}{b} \right)$Multiplying these results:$E = \left( \frac{2}{c} - \frac{1}{b} \right) \left( \frac{1}{b} \right) = \frac{2}{bc} - \frac{1}{b^2}$.

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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