Respuesta :

apologiabiology
I'm assuming you mean [tex]\frac{x^3+y^3}{x-y}[/tex]

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

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

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