show that

Solution:

Disclaimer: The question is incorrect.

$x=\frac{5-\sqrt{3}}{5+\sqrt{3}}$

$\Rightarrow x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}$

$\Rightarrow x=\frac{(5-\sqrt{3})^{2}}{5^{2}-(\sqrt{3})^{2}}$

$\Rightarrow x=\frac{25+3-10 \sqrt{3}}{25-3}$

$\Rightarrow x=\frac{28-10 \sqrt{3}}{22}$

$\Rightarrow x=\frac{14-5 \sqrt{3}}{11}$

$y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$

$\Rightarrow y=\frac{5+\sqrt{3}}{5-\sqrt{3}} \times \frac{5+\sqrt{3}}{5+\sqrt{3}}$

$\Rightarrow y=\frac{(5+\sqrt{3})^{2}}{5^{2}-(\sqrt{3})^{2}}$

$\Rightarrow y=\frac{25+3+10 \sqrt{3}}{25-3}$

$\Rightarrow y=\frac{28+10 \sqrt{3}}{22}$

$\Rightarrow y=\frac{14+5 \sqrt{3}}{11}$

$\therefore x^{2}-y^{2}$

$=\left(\frac{14-5 \sqrt{3}}{11}\right)^{2}-\left(\frac{14+5 \sqrt{3}}{11}\right)^{2}$

$=\frac{196+75-140 \sqrt{3}}{121}-\frac{196+75+140 \sqrt{3}}{121}$

$=\frac{271-140 \sqrt{3}}{121}-\frac{271+140 \sqrt{3}}{121}$

$=\frac{271-140 \sqrt{3}-271-140 \sqrt{3}}{121}$

$=\frac{-280 \sqrt{3}}{121}$

The question is incorrect. Kindly check the question.

The question should have been to show that $x-y=-\frac{10 \sqrt{3}}{11}$.

$\therefore x-y$

$=\frac{14-5 \sqrt{3}}{11}-\frac{14+5 \sqrt{3}}{11}$

$=\frac{14-5 \sqrt{3}-14-5 \sqrt{3}}{11}$

$=\frac{-10 \sqrt{3}}{11}$