Respuesta :

gmany

[tex] The\ domain:\\D:y\neq-1\\\\(y^4-1)\div(y+1)=\dfrac{(y^2)^2-1^2}{y+1}=\dfrac{(y^2-1)(y^2+1)}{y+1}\\\\=\dfrac{(y-1)(y+1)(y^2+1)}{y+1}=(y+1)(y^2+1)=y^3+y^2+y+1\\\\Used:\\(a^n)^m=a^{nm}\\\\a^2-b^2=(a-b)(a+b) [/tex]

kaitlyncha

Answer: y^3 - y^2 + y - 1


Step-by-step explanation: Factor the numerator and denominator and cancel the common factors. This should leave you with y^3-y^2+y-1.

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