Question

X^6-64=0, em complexos.

Answer 1

$x^6-64=0\\(x^3+8)\cdot(x^3-8)=0\\(x+2)(x^2-2x+4)\cdot(x-2)(x^2+2x+4)=0\\(x+2)(x-2)^2\cdot(x-2)(x+2)^2=0\\(x+2)(x-2)(x-2)\cdot(x-2)(x+2)(x+2)=0\\\boxed{S=\left\{-2,2\right\}}$