Factor (xy - 1)^2 + (x + y - 2)(x + y - 2xy) completely.

Step-by-step solution

1. 1

   Expand the square. Using $(a-b)^{2}=a^{2}-2ab+b^{2}$ with $a=xy$ and $b=1$:

   $(xy-1)^{2}=x^{2}y^{2}-2xy+1.$

2. 2

   Expand the product in two halves. Splitting the first bracket as $(x+y)-2$:

   $(x+y)(x+y-2xy)=(x+y)^{2}-2xy(x+y)=x^{2}+2xy+y^{2}-2x^{2}y-2xy^{2},$

   $-2(x+y-2xy)=-2x-2y+4xy.$

   Adding these:

   $x^{2}+y^{2}+6xy-2x^{2}y-2xy^{2}-2x-2y.$

   The $6xy$ comes from $2xy+4xy$ — collecting those two separate contributions is the step most likely to go wrong.

3. 3

   Add both parts together. Combining with the expanded square and merging the $xy$ terms ($-2xy+6xy=4xy$):

   $x^{2}y^{2}-2x^{2}y-2xy^{2}+x^{2}+y^{2}+4xy-2x-2y+1.$

   Nine terms, symmetric under swapping $x$ and $y$ — a strong hint that the factorisation will be symmetric too.

4. 4

   Recognise the perfect square. Consider $t=xy-x-y+1$. Squaring it:

   $t^{2}=x^{2}y^{2}+x^{2}+y^{2}+1-2x^{2}y-2xy^{2}+2xy+2xy-2x-2y=x^{2}y^{2}-2x^{2}y-2xy^{2}+x^{2}+y^{2}+4xy-2x-2y+1,$

   which matches the expansion exactly. So the whole expression equals $t^{2}$.

5. 5

   Factor $t$ and state the answer. Grouping, $xy-x-y+1=x(y-1)-(y-1)=(x-1)(y-1)$, so

   $(xy-1)^{2}+(x+y-2)(x+y-2xy)=\left[(x-1)(y-1)\right]^{2}=(x-1)^{2}(y-1)^{2}.$

   The expression is therefore never negative, and it vanishes exactly when $x=1$ or $y=1$. Numerical check at $(x,y)=(1.8,-0.4)$: both forms give $1.2544$ ✓, with two further random pairs agreeing to nine digits.