Question

Include parentheses in your answer. (x3 + y3) ÷ (x - y)

Answer 1

I'm assuming you mean $\frac{x^3+y^3}{x-y}$

we can factor that sum of perfect cubes from $x^2+y^3$ into $(x+y)(x^2-xy+y^2)$

so therefor $\frac{x^3+y^3}{x-y}=\frac{(x+y)(x^2-xy+y^2)}{x-y}$
that is simpliest form