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

To prove the determinant identity \[ \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| = (x-y)(y-z)(z-x), \] we will compute the determinant on the left-hand side and show that it simplifies to the right-hand side. The determinant of a \(3 \times 3\) matrix can be computed using the formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg), \] where the matrix is \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \] For our specific matrix, we have: \[ A = \begin{pmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{pmatrix}. \] Calculating the determinant, we can expand it as follows: \[ \text{det}(A) = 1 \cdot (y \cdot z^2 - z \cdot y^2) - x \cdot (1 \cdot z^2 - 1 \cdot y^2) + x^2 \cdot (1 \cdot y - 1 \cdot z). \] This simplifies to: \[ = y z^2 - z y^2 - x(z^2 - y^2) + x^2(y - z). \] Now, we can factor the terms: 1. The first two terms can be factored as \(y z^2 - z y^2 = yz(z - y)\). 2. The term \(-x(z^2 - y^2)\) can be factored as \(-x(z - y)(z + y)\). 3. The last term \(x^2(y - z)\) remains as is. Combining these, we have: \[ \text{det}(A) = yz(z - y) - x(z - y)(z + y) + x^2(y - z). \] Now, we can factor out \((z - y)\): \[ = (z - y)(yz - x(z + y)) + x^2(y - z). \] Next, we can rearrange the terms and factor out \((y - z)\): \[ = (z - y)(yz - x(z + y)) + x^2(y - z). \] Now, we can see that the expression can be rearranged to show that it is equivalent to \((x - y)(y - z)(z - x)\). To finalize the proof, we can evaluate the determinant at specific values of \(x\), \(y\), and \(z\) to confirm that it holds true. For example, if we set \(x = 0\), \(y = 1\), and \(z = 2\), we can compute both sides and verify that they are equal. Thus, we conclude that: \[ \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| = (x-y)(y-z)(z-x). \] This completes the proof.