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If $\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}$ are in A. P., then $\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right) \left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)$ is equal to
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The correct answer is:
$\frac{4}{a c}-\frac{3}{b^{2}}$
$\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}$
$\therefore\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right)\left(\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$
$=\left(\frac{2}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)\left(\frac{2}{\mathrm{c}}-\frac{1}{\mathrm{~b}}\right)=\frac{4}{\mathrm{ac}}-\frac{1}{\mathrm{~b}}\left(\frac{2}{\mathrm{a}}+\frac{2}{\mathrm{c}}\right)+\frac{1}{\mathrm{~b}^{2}}$
$=\frac{4}{\mathrm{ac}}-\frac{2}{\mathrm{~b}}\left(\frac{2}{\mathrm{~b}}\right)+\frac{1}{\mathrm{~b}^{2}}=\frac{4}{\mathrm{ac}}-\frac{3}{\mathrm{~b}^{2}}$
$\therefore\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right)\left(\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$
$=\left(\frac{2}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)\left(\frac{2}{\mathrm{c}}-\frac{1}{\mathrm{~b}}\right)=\frac{4}{\mathrm{ac}}-\frac{1}{\mathrm{~b}}\left(\frac{2}{\mathrm{a}}+\frac{2}{\mathrm{c}}\right)+\frac{1}{\mathrm{~b}^{2}}$
$=\frac{4}{\mathrm{ac}}-\frac{2}{\mathrm{~b}}\left(\frac{2}{\mathrm{~b}}\right)+\frac{1}{\mathrm{~b}^{2}}=\frac{4}{\mathrm{ac}}-\frac{3}{\mathrm{~b}^{2}}$
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