Divide: by
pllzzzz help
Factorize the numerator!
By grouping we get
$\displaystyle x^4+x^3y-xy^3-y^4 = x^3(x+y)-y^3(x+y) = (x+y)(x^3-y^3)$
This should get you started. Note the 2nd factor is the difference of cubes and can be factored as well.
after you have factored both the numerator and denominator the entire denominator should reduce out.
Try to work this out and post back with your work if you get stuck.
You can almost do this "by inspection".
I see $\displaystyle (x^4 - y^4)$, so I see an $\displaystyle (x^2 - y^2)$ in there.
The rest of the numerator is $\displaystyle x^3y-xy^3$, which factors into $\displaystyle xy(x^2 - y^2)$.
Hello, koolaid123!
$\displaystyle \text{Divide: }\:(x^4+x^3y-xy^3 - y^4) \div (x^2-y^2)$
I don't suppose you tried long division . . .
$\displaystyle \begin{array}{ccccccccccccc} &&&&&& x^2 &+& xy &+&y^2 \\ && --&--&--&--&--&--&--&--&-- \\x^2-y^2 & | & x^4 &+& x^3y && &-& xy^3 & - & y^4 \\ && x^4 && &-& x^2y^2 \\&& --&--&--&--&-- \\ &&&& x^3y &+& x^2y^2 &-& xy^3 \\ &&&& x^3y && &-& xy^3 \\ &&&& --&--&--&--&-- \\ &&&&&& x^2y^2 && &-& y^4 \\ &&&&&& x^2y^2 && &-& y^4 \\ &&&&&& --&--&--&--&-- \end{array}$