Question : If $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ then the value of $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?
Option 1: $\frac{361}{363}$
Option 2: $\frac{341}{343}$
Option 3: $\frac{384}{387}$
Option 4: $\frac{321}{323}$
Correct Answer: $\frac{321}{323}$
Solution :
$x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}} = \frac{(\sqrt{5}-\sqrt{4})^2}{(\sqrt{5}+\sqrt{4})(\sqrt{5}-\sqrt{4})} = \frac{5+4-4\sqrt5}{5-4} = 9-4\sqrt5$
$y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}} = \frac{(\sqrt{5}+\sqrt{4})^2}{(\sqrt{5}-\sqrt{4})(\sqrt{5}+\sqrt{4})} = \frac{5+4+4\sqrt5}{5-4} = 9+4\sqrt5$
⇒ $xy=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}×\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}=1$
and $x+y=9-4\sqrt5+9+4\sqrt5=18$
Now, $\frac{x^2-x y+y^2}{x^2+x y+y^2}$
= $\frac{x^2+2xy+y^2-3xy}{x^2+2xy+y^2-xy}$
= $\frac{(x+y)^2-3xy}{(x+y)^2-xy}$
= $\frac{(18)^2-3}{(18)^2-1}$
= $\frac{321}{323}$
Hence, the correct answer is $\frac{321}{323}$.
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