2.182. $\frac{\left(a b\left(x^{2}+y^{2}\right)+x y\left(a^{2}+b^{2}\right)\right) \cdot\left((a x+b y)^{2}-4 a b x y\right)}{a b\left(x^{2}-y^{2}\right)+x y\left(a^{2}-b^{2}\right)}$.
Solution.

Domain of definition: $a b\left(x^{2}-y^{2}\right)+x y\left(a^{2}-b^{2}\right) \neq 0, a b x^{2}+\left(a^{2}-b^{2}\right) y x-a b y^{2} \neq 0$,

$$
\begin{aligned}
& x_{1,2} \neq \frac{-\left(a^{2}-b^{2}\right) y \pm y \sqrt{\left(a^{2}-b^{2}\right)^{2} y^{2}+4 a^{2} b^{2} y^{2}}}{2 a b}= \\
& =\frac{-\left(a^{2}-b^{2}\right) y \pm y \sqrt{a^{4}-2 a^{2} b^{2}+b^{4}+4 a^{2} b^{2}}}{2 a b}= \\
& =\frac{-\left(a^{2}-b^{2}\right) y \pm y \sqrt{a^{4}+2 a^{2} b^{2}+b^{4}}}{2 a b}= \\
& =\frac{-\left(a^{2}-b^{2}\right) y \pm y \sqrt{\left(a^{2}+b^{2}\right)^{2}}}{2 a b}=\frac{-\left(a^{2}-b^{2}\right) y \pm\left(a^{2}+b^{2}\right) y}{2 a b}= \\
& =\frac{\left(-a^{2}+b^{2} \pm\left(a^{2}+b^{2}\right)\right) y}{2 a b} \\
& \left\{\begin{array} { l } 
{ x _ { 1 } \neq \frac { ( - a ^ { 2 } + b ^ { 2 } - ( a ^ { 2 } + b ^ { 2 } ) ) y } { 2 a b } = \frac { - 2 a ^ { 2 } y } { 2 a b } = - \frac { a y } { b } , } \\
{ x _ { 2 } \neq \frac { ( - a ^ { 2 } + b ^ { 2 } + ( a ^ { 2 } + b ^ { 2 } ) ) y } { 2 a b } = \frac { 2 b ^ { 2 } y } { 2 a b } = \frac { b y } { a } , }
\end{array} \text { i.e. } \left\{\begin{array}{l}
x \neq-\frac{a y}{b} \\
x \neq \frac{b y}{a} \\
a b \neq 0
\end{array}\right.\right. \\
& \frac{\left(a b\left(x^{2}+y^{2}\right)+x y\left(a^{2}+b^{2}\right)\right)\left((a x+b y)^{2}-4 a b x y\right)}{a b\left(x^{2}-y^{2}\right)+x y\left(a^{2}-b^{2}\right)}= \\
& =\frac{\left(a b x^{2}+\left(a^{2}+b^{2}\right) y x+a b y^{2}\right)\left(a^{2} x^{2}+2 a b x y+b^{2} y^{2}-4 a b x y\right)}{a b x^{2}+\left(a^{2}-b^{2}\right) y x-a b y^{2}}= \\
& =\frac{\left(a b x^{2}+\left(a^{2}+b^{2}\right) y x+a b y^{2}\right)(a x-b y)^{2}}{a b x^{2}+\left(a^{2}-b^{2}\right) y x-a b y^{2}}
\end{aligned}
$$

We factorize the numerator $a b x^{2}+\left(a^{2}+b^{2}\right) y x+a b y^{2}$ as a quadratic trinomial in $x$:

$$
\begin{aligned}
& x_{1,2}=\frac{-\left(a^{2}+b^{2}\right) y \pm \sqrt{\left(a^{2}+b^{2}\right) y^{2}-4 a^{2} b^{2} y^{2}}}{2 a b}= \\
& =\frac{-\left(a^{2}+b^{2}\right) y \pm y \sqrt{a^{4}+2 a^{2} b^{2}+b^{4}-4 a^{2} b^{2}}}{2 a b}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{-\left(a^{2}+b^{2}\right) y \pm y \sqrt{a^{4}-2 a^{2} b^{2}+b^{4}}}{2 a b}= \\
& =\frac{-\left(a^{2}+b^{2}\right) y \pm y \sqrt{\left(a^{2}-b^{2}\right)^{2}}}{2 a b}=\frac{-\left(a^{2}+b^{2}\right) y \pm\left(a^{2}-b^{2}\right) y}{2 a b} \\
& \text { or } x_{1}=\frac{\left(-a^{2}-b^{2}-a^{2}+b^{2}\right) y}{2 a b}=\frac{-2 a^{2} y}{2 a b}=-\frac{a y}{b} \\
& x_{2}=\frac{\left(-a^{2}-b^{2}+a^{2}-b^{2}\right) y}{2 a b}=\frac{-2 b^{2} y}{2 a b}=-\frac{b y}{a}
\end{aligned}
$$

then

$$
a b x^{2}+\left(a^{2}+b^{2}\right) y x+a b y^{2}=a b\left(x+\frac{a y}{b}\right)\left(x-\frac{b y}{a}\right)=\left(a b x+a^{2} y\right)\left(a b x-b^{2} y\right)
$$

Similarly, we transform the denominator. Then the expression (*) will take the form

$$
\begin{aligned}
& \frac{\left(a b x+a^{2} y\right)\left(a b x+b^{2} y\right)(a x-b y)^{2}}{\left(a b x+a^{2} y\right)\left(a b x-b^{2} y\right)}=\frac{\left(a b x+b^{2} y\right)(a x-b y)^{2}}{b(a x-b y)}= \\
& =\frac{b(a x+b y)(a x-b y)}{b}=a^{2} x^{2}-b^{2} y^{2}
\end{aligned}
$$

Answer: $a^{2} x^{2}-b^{2} y^{2}$.