Question

Let $1,\omega$ and ${\omega }^2$ be the cube roots of unity. The least possible degree of a polynomial with real coefficients, having $2{\omega }^2,3+4\omega ,3+4{\omega }^2$ and $5-\omega -{\omega }^2$ as roots is

Answer 1

Complex roots occur in conjugate pairs.
$\omega =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
${\omega }^2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=\frac{-1}{2}-\frac{\sqrt{3}}{2}i$
$\omega$ and ${\omega }^2$ are conjugate pairs.
Also, $1+\omega +{\omega }^2=0$

Consider the root $2{\omega }^2$.
Its conjugate is $2\omega$.

$3+4\omega$ and $3+4{\omega }^2$ are conjugate pairs.

$5-\omega -{\omega }^2=5-(-1)=6$ which is real.
$\therefore$ Polynomial has minimum of $5$ roots.