assumin you mean
[tex]\frac{x^4-1}{x+1}[/tex]
recognize difference of 2 perfect squares
remember that [tex]a^2-b^2=(a+b)(a-b)[/tex]
so
factor the numerator
[tex]x^4-1=(x^2)^2-(1)^2=[/tex] [tex] (x^2+1)(x^2-1)[/tex]
but look, another perfect square
[tex](x^2+1)(x^2-1)=(x^2+1)(x^2-1^2)=(x^2+1)(x+1)(x-1)[/tex]
so we end up with
[tex]\frac{x^4-1}{x+1}=\frac{(x^2+1)(x-1)(x+1)}{x+1}[/tex] which simplifies to [tex](x^2+1)(x-1)[/tex]
