Respuesta :

LammettHash

Factorize the numerator as a difference of squares:

[tex]y^4-1=(y^2-1)(y^2+1)=(y-1)(y+1)(y^2+1)[/tex]

Then if [tex]y\neq-1[/tex], the [tex]y+1[/tex] factors will cancel:

[tex]\dfrac{y^4-1}{y+1}=(y-1)(y^2+1)=y^3-y^2+y-1[/tex]

so the answer is B.

Answer is B.) y3 - y2 + y - 1