If $\omega$ is a complex cube root of unity, show that the following equals null matrix. 
If $\omega$ is a complex cube root of unity, show that
  $$ \left(  \begin{bmatrix}
        1 & \omega & \omega^2 \\
       \omega & \omega^2 & 1 \\
        \omega^2 & 1 & \omega \\
        \end{bmatrix} + \begin{bmatrix}
        \omega & \omega^2 & 1 \\
        \omega^2 & 1 & \omega \\
        \omega & \omega^2 & 1 \\
        \end{bmatrix} \right) 
     \begin{bmatrix}
        1  \\
        \omega  \\
        \omega^2  \\
        \end{bmatrix}=\begin{bmatrix}
        0  \\
        0  \\
        0  \\
        \end{bmatrix}$$

I tried to solve this and I reduced the L.H.S. to
$$\begin{bmatrix}
        2 +2\omega+2\omega^2 \\
        2 +2\omega+2\omega^2 \\
        2 +2\omega+2\omega^2 \\
        \end{bmatrix} 
$$ (since $\omega^3=1$)
but couldn't equate it to R.H.S.
Please provide your assistance.
Thank you

$$\omega^3=1 \Rightarrow ω^3−1=0 \Rightarrow (ω−1)(ω^2+ω+1)=0 \Rightarrow ω−1=0 \lor  ω^2+ω+1= 0$$
Since, $\omega \neq 1 $ as it is complex , $ω^2+ω+1=0$
Hence, LHS = RHS
- Answer originally posted as comment by Git Gud
