• # question_answer If $\omega$ is a complex cube root of unity, then $\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}+\frac{c+a\omega +b{{\omega }^{2}}}{a+b\omega +c{{\omega }^{2}}}+\frac{b+c\omega +a{{\omega }^{2}}}{b+c{{\omega }^{4}}+a{{\omega }^{5}}}$ is equal to: A)  $1$                    B)         $\omega$                      C)  ${{\omega }^{2}}$                       D)         $-1$E)  0

$\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}+\frac{c+a\omega +b{{\omega }^{2}}}{a+b\omega +c{{\omega }^{2}}}+\frac{b+c\omega +a{{\omega }^{2}}}{b+c{{\omega }^{4}}+a{{\omega }^{5}}}$ $=\frac{{{\omega }^{2}}(a+b\omega +c{{\omega }^{2}})}{(a+b\omega +c{{\omega }^{2}})}+\frac{\omega (a\omega +b{{\omega }^{2}}+c)}{(a\omega +b{{\omega }^{2}}+c)}$                                                 $+\frac{b+c\omega +a{{\omega }^{2}}}{b+c\omega +a{{\omega }^{2}}}$ $={{\omega }^{2}}+\omega +1=0$