What are the cube roots of 16?

1 Answer
Apr 4, 2017

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

Explanation:

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