Prove 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 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 of the matrix on the left-hand side. Let \[ D = \left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|. \] We can use the properties of determinants to simplify this calculation. We can perform row operations without changing the value of the determinant. Specifically, we can subtract the first row from the second and third rows: \[ D = \left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & y-x & y^{2}-x^{2} \\ 0 & z-x & z^{2}-x^{2} \end{array}\right|. \] Next, we simplify the second and third rows. Notice that \(y^2 - x^2\) and \(z^2 - x^2\) can be factored: \[ y^2 - x^2 = (y-x)(y+x), \] \[ z^2 - x^2 = (z-x)(z+x). \] Thus, we can rewrite the determinant as: \[ D = \left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & y-x & (y-x)(y+x) \\ 0 & z-x & (z-x)(z+x) \end{array}\right|. \] Now, we can factor out \(y-x\) from the second row and \(z-x\) from the third row: \[ D = (y-x)(z-x) \left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & 1 & y+x \\ 0 & 1 & z+x \end{array}\right|. \] Next, we can simplify the determinant of the \(2 \times 2\) matrix: \[ \left|\begin{array}{cc} 1 & y+x \\ 1 & z+x \end{array}\right| = (z+x) - (y+x) = z - y. \] Putting it all together, we have: \[ D = (y-x)(z-x)(z-y). \] Now, we can rearrange the factors to match the desired form: \[ D = (x-y)(y-z)(z-x). \] Thus, we have shown 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. \[ \boxed{(x-y)(y-z)(z-x)}. \]