What are the cube roots of $16$ ?

Question

2146 views around the world

You can reuse this answer
Creative Commons License

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 $k=root(3)16$

What are the cube roots of 1616?