Maths- Calculus

By the straight application of de Moivre's formula :

z^(1/n)
= (r(cos(x)+i sin(x)))^(1/r)
= r^(1/n)(cos((x+2kπ)/n)+i sin((x+2kπ )/n))

For z=-8, we get
r=2,
x= π, (corresponding to the point (-1, 0)) for which the roots are
π/3, (π+2π)/3 or (π+4π)/3
= π/3, π , 5π /3 corresponding to 60, 180 and 300 degrees
Thus the roots of -8 are
2(cos(π/3), i sin(π/3)) , 2(cos(π), i sin(π)) , and 2(cos(5π/3), i sin(5π/3))
=2(1/2, i sqrt(3)/2), 2(-1, 0) , and 2(1/2, -i sqrt(3)/2)
=1+sqrt(3) i, -2, and 1-sqrt(3) i

z^3=-8

z^3 +8=0

(z+2)(z^2-2z+4)=0

z= -2 or z=1+√3i or z=1-√3i