1. ## Is this accurate?

$\displaystyle (x+y+xy)^2$

Solved:

$\displaystyle (x+y+xy)(x+y+xy)$

= $\displaystyle x(x+y+xy)+y(x+y+xy)+xy(x+y+xy)$

= $\displaystyle x^2+xy+x^2y+xy+y^2+xy^2+x^2y+xy^2+x^2y^2$

=$\displaystyle x^2y^2+x^2+y^2+2xy+2x^2y+2xy^2+x^2y$

2. Hello, deekay930!

Small slip at the end . . .

$\displaystyle (x+y+xy)^2$

Solved:

$\displaystyle (x+y+xy)(x+y+xy)$

. . $\displaystyle =\;x(x+y+xy)+y(x+y+xy)+xy(x+y+xy)$

. . $\displaystyle \;=x^2+xy+ {\color{blue}x^2y}+xy+y^2+xy^2+ {\color{blue}x^2y}+xy^2+x^2y^2$

. . $\displaystyle =\;x^2y^2 + x^2 + y^2 + 2xy + {\color{blue}2x^2y} + 2xy^2 +\:{\color{red}\rlap{////}}x^2y$

3. Originally Posted by deekay930
$\displaystyle (x+y+xy)^2$

Solved:

$\displaystyle (x+y+xy)(x+y+xy)$

= $\displaystyle x(x+y+xy)+y(x+y+xy)+xy(x+y+xy)$

= $\displaystyle x^2+xy+x^2y+xy+y^2+xy^2+x^2y+xy^2+x^2y^2$

=$\displaystyle x^2y^2+x^2+y^2+2xy+2x^2y+2xy^2+x^2y$
Yes. this is accurate. Except for the extra term...