Find roots of polynomial $$ p(x) = x^{20}-3 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 1.0565\\[1 em]x_2 &= -1.0565\\[1 em]x_3 &= -0.621+0.8547i\\[1 em]x_4 &= -0.621-0.8547i\\[1 em]x_5 &= -0.3265+1.0048i\\[1 em]x_6 &= -0.3265-1.0048i\\[1 em]x_7 &= 0.621+0.8547i\\[1 em]x_8 &= 0.621-0.8547i\\[1 em]x_9 &= 0.3265+1.0048i\\[1 em]x_{10} &= 0.3265-1.0048i\\[1 em]x_{11} &= -0.8547+0.621i\\[1 em]x_{12} &= -0.8547-0.621i\\[1 em]x_{13} &= 1.0565i\\[1 em]x_{14} &= -1.0565i\\[1 em]x_{15} &= 0.8547+0.621i\\[1 em]x_{16} &= 0.8547-0.621i\\[1 em]x_{17} &= -1.0048+0.3265i\\[1 em]x_{18} &= -1.0048-0.3265i\\[1 em]x_{19} &= 1.0048+0.3265i\\[1 em]x_{20} &= 1.0048-0.3265i \end{aligned} $$

Polynomial $ x^{20}-3 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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