Respuesta :
[tex]y=1+i\qquad\qquad i^2=-1\\\\\\
y^3-3y^2+(?)-1 =-i\\\\(1+i)^3-3\cdot(1+i)^2+(?)-1+i=0\\\\1^3+3i+3i^2+i^3-3\cdot(1^2+2i+i^2)+(?)-1+i=0\\\\
1+3i+3\cdot(-1)+i\cdot i^2-3-6i-3i^2+(?)-1+i=0\\\\
1+3i-3+i\cdot(-1)-3-6i-3\cdot(-1)+(?)-1+i=0\\\\
1+3i-3-i-3-6i+3+(?)-1+i=0\\\\
-3-3i+(?)=0\\\\(?)=3+3i\\\\(?)=3(1+i)\\\\\boxed{(?)=3y}[/tex]