How do you find the cube roots of $-125$ ?

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Answer 1 · 2016-09-08T09:49:31

$5(1/2 + (i sqrt[3])/2),-5,5(1/2 - (i sqrt[3])/2)$

Using the de Moivre's identity

$-125 = 5^3 e^(i(pi+2kpi)), k=0,pm1,pm2,cdots$ because

$e^(i(pi+2kpi)) = cos(pi+2kpi)+i sin(pi+2kpi) = -1$

so

$(-125)^(1/3) = (5^3 e^(i(pi+2kpi)))^(1/3) = 5e^(i(pi+2kpi)/3)$

for $k=0, k=pm1$ we have

$5e^(ipi/3),5e^(ipi),5e^(-ipi/3)$

or

$5(1/2 + (i sqrt[3])/2),-5,5(1/2 - (i sqrt[3])/2)$

How do you find the cube roots of -125-125?