Find the value of x and y using the equations: ${\mathrm{x}}^4+{\mathrm{y}}^4=82$ and $\mathrm{x}-\mathrm{y}=2$.

Given equations:

${\mathrm{x}}^4+{\mathrm{y}}^4=82$ ………… (1)

$\mathrm{x}-\mathrm{y}=2$ ………….. (2)

Now, we square equation (2).

So, we obtain:

$(\mathrm{x}-\mathrm{y}{)}^2=4$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}=4$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2 = 4+ 2\mathrm{xy}$

Squaring both sides:

$\Rightarrow {\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2=(4+2\mathrm{xy}{)}^2$

$\Rightarrow {\mathrm{x}}^4+{\mathrm{y}}^4+2{\mathrm{x}}^2{\mathrm{y}}^2=16+4{\mathrm{x}}^2{\mathrm{y}}^2+16\mathrm{xy}$ ………….(3)

Again, we take the value of equation (1) in equation (3).

So, we obtain:

$82=16+2{\mathrm{x}}^2{\mathrm{y}}^2+16\mathrm{xy}$

$\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2+8\mathrm{xy}=33$

$\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2+8\mathrm{xy}-33=0$

Now, we factorize the equation:

$\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2+8\mathrm{xy}-33=0$

$\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2+11\mathrm{xy}-3\mathrm{xy}-33=0$

$\Rightarrow \mathrm{xy}\left(\mathrm{xy}+11\right)-3\left(\mathrm{xy}+11\right)=0$

$\Rightarrow \left(\mathrm{xy}-3\right)\left(\mathrm{xy}+11\right)=0$

$\mathrm{xy}=3\text{ or }\mathrm{xy}=-11$

From equation (2),

$\mathrm{x}-\mathrm{y}=2$

$\Rightarrow \mathrm{x}=\mathrm{y}+2$ ……………(4)

Put the value $\mathrm{xy}=3$, i.e. $x= \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$y$}\right.$

$\Rightarrow \mathrm{y}(\mathrm{y}+2)=3$

$\Rightarrow {\mathrm{y}}^2+2\mathrm{y}-3=0$

$\Rightarrow {\mathrm{y}}^2+3\mathrm{y}-\mathrm{y}-3=0$

$\Rightarrow \mathrm{y}\left(\mathrm{y}+3\right)-1\left(\mathrm{y}+3\right)=0$

$\Rightarrow \left(\mathrm{y}-1\right)\left(\mathrm{y}+3\right)=0$

$\Rightarrow \mathrm{y}=1\text{ or }\mathrm{y}=-3$

Also, we get:

$\mathrm{x}=\mathrm{y}+2=1+2=3$

And $\mathrm{x}=\mathrm{y}+2=-3+2=-1$

Put the value $\mathrm{xy}=-11$

$\Rightarrow \mathrm{y}\left(\mathrm{y}+2\right)=-11$

$\Rightarrow {\mathrm{y}}^2+2\mathrm{y}+11=0$

Discriminant = $\sqrt{4-44}$ 

This will have no real roots as the discriminant is less than zero. 

Hence, the required value of given equation is:$\mathrm{y}=1\text{ or }\mathrm{y}=-3$ and $\mathrm{x}=3\text{ or }\mathrm{x}=-1$.