$\mathbf{\omega }$ is a complex cube root of unity. When an unbiased die is thrown 3 times, if  ${\mathbf{\beta }}_1\mathbf{,}{\mathbf{\beta }}_2$ and  ${\mathbf{\beta }}_3$ are the numbers appeared on the die, then the probability that ${\mathbf{\beta }}_1\mathbf{,}{\mathbf{\beta }}_2$ and ${\mathbf{\beta }}_3$ satisfy ${\mathbf{\omega }}^{{\mathbf{\beta }}_1}\mathbf{+}{\mathbf{\omega }}^{{\mathbf{\beta }}_2}\mathbf{+}{\mathbf{\omega }}^{{\mathbf{\beta }}_3}\mathbf{=}\mathbf{0}$ is

n(s) = 216

${\text{ω}}^{{\text{β}}_{\text{1}}}{\text{+\hspace{0.17em}ω}}^{{\text{β}}_{\text{2}}}{\text{\hspace{0.17em}+\hspace{0.17em}ω}}^{{\text{β}}_{\text{3}}}\text{=0}$

$\left({\text{β}}_{\text{1}}{\text{,\hspace{0.17em}\hspace{0.17em}β}}_{\text{2}}{\text{,\hspace{0.17em}\hspace{0.17em}β}}_{\text{3}}\right)$

(1, 2, 3) $\stackrel{ }{\to }$ 6 ways

(2, 3, 4) $\stackrel{ }{\to }$ 6 ways

(3, 4, 5) $\stackrel{ }{\to }$ 6ways

(4, 5, 6) $\stackrel{ }{\to }$ 6 ways

(1, 3, 5) $\stackrel{ }{\to }$ 6 ways

(2, 4, 6) $\stackrel{ }{\to }$ 6 ways

(1, 5, 6) $\stackrel{ }{\to }$ 6 ways

(1, 2, 6) $\stackrel{ }{\to }$ 6 ways

                     48

Prob $\text{=\hspace{0.17em}}\frac{\text{48}}{\text{216}}\text{=}\frac{\text{2}}{\text{9}}$