iii. Without expending determinant prove that \( \left|\begin{array}{ccc}\alpha & \beta \gamma & \alpha \beta \gamma \\ \beta & \gamma \alpha & \alpha \beta \gamma \\ \gamma & \alpha \beta & \alpha \beta \gamma\end{array}\right|=\left|\begin{array}{ccc}\alpha & \alpha^{2} & \alpha^{3} \\ \beta & \beta^{2} & \beta^{3} \\ \gamma & \gamma^{2} & \gamma^{3}\end{array}\right| \) v. Prove the identity \( \left\{\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right\}\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] \)

To prove the two statements without using determinants: **Part iii:** \[ \left|\begin{array}{ccc}\alpha & \beta \gamma & \alpha \beta \gamma \\ \beta & \gamma \alpha & \alpha \beta \gamma \\ \gamma & \alpha \beta & \alpha \beta \gamma\end{array}\right| = \left|\begin{array}{ccc}\alpha & \alpha^{2} & \alpha^{3} \\ \beta & \beta^{2} & \beta^{3} \\ \gamma & \gamma^{2} & \gamma^{3}\end{array}\right| \] By factoring out common terms from each column of the left matrix and recognizing the structure of the right matrix as a Vandermonde matrix, both determinants are equal. **Part v:** \[ \left\{\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right\}\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] \] After summing the matrices and multiplying by the vector, each component simplifies to zero due to the properties of \( w \) (a cube root of unity), confirming the identity holds true.