應該不用幾分鐘
\(\begin{align}
  & {{\omega }^{7}}=1,\omega \ne 1 \\
& {{\omega }^{6}}+{{\omega }^{5}}+{{\omega }^{4}}+{{\omega }^{3}}+{{\omega }^{2}}+\omega =-1 \\
&  \\
& \alpha +\beta +\gamma =\omega +{{\omega }^{6}}+{{\omega }^{2}}+{{\omega }^{5}}+{{\omega }^{3}}+{{\omega }^{4}}=-1 \\
& \alpha \beta +\beta \gamma +\gamma \alpha =\left( \omega +{{\omega }^{6}} \right)\left( {{\omega }^{2}}+{{\omega }^{5}} \right)+\left( {{\omega }^{2}}+{{\omega }^{5}} \right)\left( {{\omega }^{3}}+{{\omega }^{4}} \right)+\left( {{\omega }^{3}}+{{\omega }^{4}} \right)\left( \omega +{{\omega }^{6}} \right) \\
& =2\left( {{\omega }^{6}}+{{\omega }^{5}}+{{\omega }^{4}}+{{\omega }^{3}}+{{\omega }^{2}}+\omega  \right)=-2 \\
& \alpha \beta \gamma =\left( \omega +{{\omega }^{6}} \right)\left( {{\omega }^{2}}+{{\omega }^{5}} \right)\left( {{\omega }^{3}}+{{\omega }^{4}} \right)=2+\left( {{\omega }^{6}}+{{\omega }^{5}}+{{\omega }^{4}}+{{\omega }^{3}}+{{\omega }^{2}}+\omega  \right)=1 \\
\end{align}\)

所求為\({{x}^{3}}+{{x}^{2}}-2x-1=0\)

同一題，換個樣子罷了