Răspuns :
z³+8=0
z³+2³=0
(z+2)(z²-2z+4)=0
z+2=0 z=-2
z²-2z+4=0
Δ=(-2)²-4×1×4=4-16=-12 i²=-1 Δ=12i²
z₁=(2-√12i²)/2=(2-2i√3)/2=1-i√3
z₂=(2+√12i²)/2=(2+2i√3)/2=1+i√3
z³+2³=0
(z+2)(z²-2z+4)=0
z+2=0 z=-2
z²-2z+4=0
Δ=(-2)²-4×1×4=4-16=-12 i²=-1 Δ=12i²
z₁=(2-√12i²)/2=(2-2i√3)/2=1-i√3
z₂=(2+√12i²)/2=(2+2i√3)/2=1+i√3