Question

Sol. [Key Concept: $a,b,c$ in H.P. ⇒ $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ in A.P.; substitute to simplify the expression.]

$$\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$$

$$\begin{matrix} \left( \frac{1}{b} - \frac{1}{a} + \frac{1}{c} \right)\left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right) & \ = \left( \frac{1}{b} - \frac{1}{a} + \frac{2}{b} - \frac{1}{a} \right)\left( \frac{2}{b} - \frac{1}{a} + \frac{1}{a} - \frac{1}{b} \right) \\ & \ = \left( \frac{3}{b} - \frac{2}{a} \right)\left( \frac{1}{b} \right) = \frac{3}{b^{2}} - \frac{2}{ab} \end{matrix}$$