Question

Quais são as raízes da equação X ao cubo-8=0

Answer 1

x³ - 8 = 0 => x³ = 8 => x = ∛8

Seja Z = 8 => |Z| = 8
cosα = a/|Z| => cosα = 8/8 = 1
senα = b/|Z| => senα = 0

Logo, α = 0

$Z_{k} = \sqrt[n]{|Z|}(cos \frac{ \alpha +2k \pi }{n} +isen \frac{ \alpha +2k \pi }{n} ) \\ \\ p/k=0=\ \textgreater \ Z_{0} = \sqrt[3]{|8|} (cos \frac{0+2.0. \pi }{3} +isen \frac{0+2.0. \pi }{3} )=\ \textgreater \ \\ \\ k_{0}=2(cos0+isen0)=\ \textgreater \ k_{0}=2(1+i.0)=\ \textgreater \ k_{0}=2 \\ \\ p/k=1=\ \textgreater \ k_{1} =2(cos \frac{0+2.1. \pi }{3}+isen \frac{0+2.1. \pi }{3}) \\ \\ k_{1} =2(cos \frac{2 \pi }{3} + \frac{2 \pi }{3)} =\ \textgreater \ k_{1}=2(- \frac{1}{2} +i. \frac{ \sqrt{3} }{2} )=\ \textgreater \ k_{1}=-1+ \sqrt{3}i$

p/k=2=>$k_{2}= 2(cos \frac{0+2.2. \pi }{3} +isen \frac{0+2.2. \pi }{3 })=\ \textgreater \ k_{2} =2(cos \frac{4 \pi }{3} +isen \frac{4 \pi }{3}) \\ \\ k_{2}=2(- \frac{1}{2}- i.\frac{ \sqrt{3} }{2} )=\ \textgreater \ k_{2} =-1- \sqrt{3} i$

S = {2, -1-√3i, -1+√3i}