How do you find three cube roots of #1#?

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Answer 1 · 2017-06-26T07:42:56

Three cube roots of #1# are

#1#, #(-1-isqrt3)/2# and #(-1+isqrt3)/2#

Let #x=root(3)1#, then #x^3=1# or

#x^3-1=0#

or #(x-1)(x^2+x+1)=0#

Hence either #x=1#

or #x^2+x+1=0# and using quadratic formula

#x=(-1+-sqrt(1-4xx1xx1))/2#

= #(-1+-sqrt(-3))/2#

= #(-1+-isqrt3)/2#

Hence three cube roots of #1# are

#1#, #(-1-isqrt3)/2# and #(-1+isqrt3)/2#