Simplify the following:
$(\frac{\sqrt{3}}{\sqrt{2}+1})^2 + (\frac{\sqrt{3}}{\sqrt{2}-1})^2 +(\frac{\sqrt{2}}{\sqrt{3}})^2$

Problem Statement

Simplify the following:

$(\frac{\sqrt{3}}{\sqrt{2}+1})^2 + (\frac{\sqrt{3}}{\sqrt{2}-1})^2 +(\frac{\sqrt{2}}{\sqrt{3}})^2$

Solution

Given :

The given expression is $(\frac{\sqrt{3}}{\sqrt{2}+1})^2 + (\frac{\sqrt{3}}{\sqrt{2}-1})^2 +(\frac{\sqrt{2}}{\sqrt{3}})^2$.

To do :

We have to simplify the given expression.

Solution :

$(\frac{\sqrt{3}}{\sqrt{2}+1})^2 + (\frac{\sqrt{3}}{\sqrt{2}-1})^2 +(\frac{\sqrt{2}}{\sqrt{3}})^2$

We know that,

$(\frac{a}{b})^2 = \frac{a^2}{b^2}$ and $\sqrt{a}^2 = a$.

$\Rightarrow \frac{\sqrt{3}^2}{(\sqrt{2}+1)^2} + \frac{\sqrt{3}^2}{(\sqrt{2}-1)^2} +\frac{\sqrt{2}^2}{\sqrt{3}^2}$

$\Rightarrow \frac{3}{\sqrt{2}^2 + 1^2 + 2.\sqrt{2}.1} + \frac{3}{\sqrt{2}^2 + 1^2 - 2.\sqrt{2}.1} + \frac{2}{3}$                            $[(a+b^2)= a^2 + b^2+2ab]$

$\Rightarrow \frac{3}{2 + 1 + 2.\sqrt{2}}+\frac{3}{2 + 1 - 2.\sqrt{2}}+  \frac{2}{3}$

$\Rightarrow \frac{3}{3 + 2\sqrt{2}}+\frac{3}{3 - 2\sqrt{2}}+  \frac{2}{3}$

$\Rightarrow \frac{3(3 - 2\sqrt{2})+3(3 + 2\sqrt{2})}{3^2 - (2\sqrt{2})^2}+  \frac{2}{3}$

$\Rightarrow \frac{9 - 6\sqrt{2}+9 + 6\sqrt{2}}{9 - 8}+  \frac{2}{3}$

$\Rightarrow \frac{18}{1} + \frac{2}{3}$

$\Rightarrow \frac{18 \times 3 + 2\times 1}{3}$

$\Rightarrow \frac{54+2}{3}$

$\Rightarrow \frac {56}{3}$

Therefore, the value of $(\frac{\sqrt{3}}{\sqrt{2}+1})^2 + (\frac{\sqrt{3}}{\sqrt{2}-1})^2 +(\frac{\sqrt{2}}{\sqrt{3}})^2$ is $\frac {56}{3}$.

Tutorialspoint

Updated on: 10-Oct-2022

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