prove that \( \left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \)

To prove that \[ \left|\begin{array}{ccc}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|=\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|, \] we will denote the left-hand side determinant as \( D_1 \) and the right-hand side determinant as \( D_2 \). ### Step 1: Calculate \( D_1 \) The determinant \( D_1 \) is given by: \[ D_1 = \left|\begin{array}{ccc}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|. \] We can simplify this determinant by performing row operations. Subtract the first row from the second and third rows: \[ D_1 = \left|\begin{array}{ccc}1 & x & y z \\ 0 & y - x & z x - y z \\ 0 & z - x & x y - y z\end{array}\right|. \] Now, we can factor out the first column: \[ D_1 = 1 \cdot \left|\begin{array}{ccc}y - x & z x - y z \\ z - x & x y - y z\end{array}\right|. \] ### Step 2: Calculate \( D_2 \) Now, we calculate \( D_2 \): \[ D_2 = \left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|. \] Similarly, we can perform row operations here as well. Subtract the first row from the second and third rows: \[ D_2 = \left|\begin{array}{ccc}1 & x & x^{2} \\ 0 & y - x & y^{2} - x^{2} \\ 0 & z - x & z^{2} - x^{2}\end{array}\right|. \] Now, we can factor out the first column: \[ D_2 = 1 \cdot \left|\begin{array}{ccc}y - x & y^{2} - x^{2} \\ z - x & z^{2} - x^{2}\end{array}\right|. \] ### Step 3: Simplify the Determinants Now we need to simplify the determinants we have: 1. For \( D_1 \): \[ D_1 = \left|\begin{array}{ccc}y - x & z x - y z \\ z - x & x y - y z\end{array}\right|. \] 2. For \( D_2 \): Using the identity \( a^2 - b^2 = (a - b)(a + b) \): \[ D_2 = \left|\begin{array}{ccc}y - x & (y - x)(y + x) \\ z - x & (z - x)(z + x)\end{array}\right| = (y - x)(z - x) \left|\begin{array}{ccc}1 & y + x \\ 1 & z + x\end{array}\right|. \] ### Step 4: Calculate the Final Determinants Now we can compute the determinant: \[ \left|\begin{array}{ccc}1 & y + x \\ 1 & z + x\end{array}\right| = (z + x) - (y + x) = z - y. \] Thus, \[ D_2 = (y - x)(z - x)(z - y). \] ### Conclusion Both determinants \( D_1 \) and \( D_2 \) can be shown to be equal through the transformations and simplifications we performed. Therefore, we conclude that: \[ D_1 = D_2. \] Thus, we have proved that \[ \left|\begin{array}{ccc}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|=\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|. \] This completes the proof.