Answer :

Procedure

[tex]\begin{gathered} \frac{y^3-1}{y+4} \\ \frac{y^3}{y+4}-\frac{1}{y+4} \end{gathered}[/tex]

y^3/(y+4) - 1/(y+4)

Alternate forms

[tex]y^2-4y+16-\frac{65}{y+4}[/tex]

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