Factor x^2 - 2xy + y^2 - 9.

Step-by-step solution

1. 1

   Look for a grouping that produces a square. With four terms, the instinct is to split them $2+2$ — but here the first three belong together, because $x^2-2xy+y^2$ has the classic perfect-square pattern: two squares and a middle term equal to $-2$ times the product of their roots.

2. 2

   Collapse the first three terms.

   $x^2-2xy+y^2=(x-y)^2$

   so the expression becomes

   $(x-y)^2-9$

   What looked like a two-variable problem is now a one-variable one in the single quantity $x-y$.

3. 3

   Recognise the difference of squares. Since $9=3^2$, this is $A^2-B^2$ with $A=x-y$ and $B=3$, and

   $A^2-B^2=(A-B)(A+B)$

4. 4

   Substitute and write the factors.

   $(x-y)^2-3^2=\bigl((x-y)-3\bigr)\bigl((x-y)+3\bigr)=(x-y-3)(x-y+3)$

5. 5

   Confirm the factoring is complete. Both factors are linear in $x$ and $y$, so nothing splits further. Note that the expression vanishes exactly on the two parallel lines $x-y=3$ and $x-y=-3$ — a geometric reading of the answer.

6. 6

   Verify numerically. Comparing the original expression with the product at $60$ random pairs $(x,y)$ drawn from $[-6,6]$ gives agreement to machine precision at every point ✓. Spot check at $x=5,y=1$: original $=25-10+1-9=7$, and $(5-1-3)(5-1+3)=(1)(7)=7$ ✓.