What are the cube roots of #16#?

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Answer 1 · 2017-04-04T13:31:13

#x=root(3)16# or #(-root(3)16+iroot(3)16sqrt3)/2# or #(-root(3)16-iroot(3)16sqrt3)/2#

#x^3=16=>x^3-16=0# i.e.

#x^3-(root(3)16)^3=0#, now let #root(3)16=k#, then we have

#x^3-k^3=0#

or #(x-k)(x^2+kx+k^2)=0#

Hence #x-k=0# i.e. #x=k# i.e. #x=root(3)16#

or #x^2+kx+k^2=0# i.e. #x=(-k+-sqrt(k^2-4k^2))/2#

i.e. #x=(-k+-iksqrt3)/2=k((-1+-isqrt3)/2)#

i.e. #x=root(3)16((-1+-isqrt3)/2)#

i.e. #x=(-root(3)16+iroot(3)16sqrt3)/2# or #x=(-root(3)16-iroot(3)16sqrt3)/2#

Note - Observe that if #1,omega,omega^2# are cube roots of #1#,

cube roots of #16# are #k, komega,komega^2#, where #k=root(3)16#