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)#