if x/y +y/x = 2. find x¹² - y^12.

Please find below the solution to the asked query :

$$\begin{gathered} \frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}=2 \\ \frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{\mathrm{xy}}=2 \\ {\mathrm{x}}^2+{\mathrm{y}}^2=2\mathrm{xy} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}=0 \\ {\left(\mathrm{x}-\mathrm{y}\right)}^2=0 \\ \mathrm{x}-\mathrm{y}=0 \\ {\mathrm{x}}^{12}-{\mathrm{y}}^{12}=\left({\mathrm{x}}^6-{\mathrm{y}}^6\right)\left({\mathrm{x}}^6+{\mathrm{y}}^6\right) \left(\mathrm{Since} {\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)\right) \\ =\left({\mathrm{x}}^3-{\mathrm{y}}^3\right)\left({\mathrm{x}}^3+{\mathrm{y}}^3\right)\left({\mathrm{x}}^6+{\mathrm{y}}^6\right) \\ =\left(\mathrm{x}-\mathrm{y}\right)\left({\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2\right)\left({\mathrm{x}}^3+{\mathrm{y}}^3\right)\left({\mathrm{x}}^6+{\mathrm{y}}^6\right) \\ =0 \mathrm{ANS}... \left(\mathrm{Since} \left(\mathrm{x}-\mathrm{y}\right)=0\right) \end{gathered}$$


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