What is the cube root of #-0.000125# ?

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Answer 1 · 2017-03-03T19:56:59

#root(3)(-0.000125) = -0.05#

#-0.000125 = -125*10^(-6) = -5^3*10^(-6)#

This has one Real cube root and two complex ones.

#root(3)(-5^3*10^(-6)) = -5*10^(-2) = -0.05#

The two complex cube roots are:

#-0.05omega" "# and #" "-0.05omega^2#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

#omega# and #omega^2# are the zeros of:

#(x^3-1)/(x-1) = x^2+x+1#