Verify the following expression (3 p 7 +7 6 p )^2- (3 7 p-7 6 p )… - Vidyadip

Question

Verify the following expression
$
\left(\frac{3 p}{7}+\frac{7}{6 p}\right)^2-\left(\frac{3}{7} p-\frac{7}{6 p}\right)^2=2
$

✓

Answer

$
\begin{aligned}
\text { LHS }= & \left(\frac{3 p}{7}+\frac{7}{6 p}\right)^2-\left(\frac{3 p}{7}-\frac{7}{6 p}\right)^2 \\
= & {\left[\left(\frac{3 p}{7}\right)^2+\left(\frac{7}{6 p}\right)^2+2 \times\left(\frac{3 p}{7}\right) \times\left(\frac{7}{6 p}\right)\right] } \\
& -\left[\left(\frac{3 p}{7}\right)^2+\left(\frac{7}{6 p}\right)^2-2 \times\left(\frac{3 p}{7}\right) \times\left(\frac{7}{6 p}\right)\right]
\end{aligned}
$
[using Identities I in I term and Identity II in II term,
$
\text { where } \left.a=\frac{3 p}{7}, b=\frac{7}{6 p}\right]
$
$
\begin{array}{l}
=\frac{9 p^2}{49}+\frac{49}{36 p^2}+1-\left[\frac{9 p^2}{49}+\frac{49}{36 p^2}-1\right] \\
=\frac{9 p^2}{49}+\frac{49}{36 p^2}+1-\frac{9 p^2}{49}-\frac{49}{36 p^2}+1=2 \\
\text { RHS }=2 \\
\text { LHS }=\text { RHS } \qquad \text {Hence proved.}
\end{array}
$

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