If a, b, and c are in H.P., then the value of (ac+ab−bc)(ab+bc−ac)(abc)2 is

A
$\frac{(a + c) (3 a - c)}{4 a^{2} c^{2}}$
B
$\frac{2}{bc} - \frac{1}{b^{2}}$
C
$\frac{2}{bc} - \frac{1}{a^{2}}$
D
$\frac{(a - c) (3 a + c)}{4 a^{2} c^{2}}$

Solution:

$\begin{array}{l} (\frac{1}{b} + \frac{1}{c} - \frac{1}{a}) (\frac{1}{c} + \frac{1}{a} - \frac{1}{b}) = (\frac{1}{b} + \frac{1}{c} - \frac{2}{b} + \frac{1}{c}) (\frac{1}{c} + \frac{1}{b} - \frac{1}{c}) \\ = (\frac{2}{c} - \frac{1}{b}) \frac{1}{b} = \frac{2}{bc} - \frac{1}{b^{2}} \end{array}$
Also by eliminating b, we get the given expression
$\frac{(a + c) (3 a - c)}{4 a^{2} c^{2}}$

If a, b, and c are in H.P., then the value of $\frac{(ac + ab - bc) (ab + bc - ac)}{(abc)^{2}}$ is

Solution:

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