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

To solve this problem, we need to expand the determinant of the given matrix and show that it equals the given expression. Let's start by expanding the determinant: The determinant of a 3x3 matrix is given by: \[ \left| \begin{array} { l l l } { a } & { b } & { c } \\ { d } & { e } & { f } \\ { g } & { h } & { i } \end{array} \right| = a(ei - fh) - b(di - fg) + c(dh - eg) \] Applying this formula to our matrix, we get: \[ \left| \begin{array} { l l l } { 1 } & { x } & { x ^ { 2 } } \\ { 1 } & { y } & { y ^ { 2 } } \\ { 1 } & { z } & { z ^ { 2 } } \end{array} \right| = 1(yz^2 - y^2z) - x(1z^2 - z^2) + x^2(1y - y) \] Simplifying each term, we have: \[ = yz^2 - y^2z - xz^2 + x^2z + x^2y - x^2y \] Notice that the terms \( x^2y - x^2y \) cancel each other out, so we are left with: \[ = yz^2 - y^2z - xz^2 + x^2z \] Now, let's factor out \( z \) from the first two terms and \( x \) from the last two terms: \[ = z(yz - y^2) - x(xz - x^2) \] \[ = z(y(z - y)) - x(x(z - x)) \] \[ = z(y - z)(z - y) - x(x - z)(z - x) \] Now, we can see that we have a common factor of \( (z - y) \) and \( (x - z) \) in both terms. We can factor these out: \[ = (z - y)(z - y) - (x - z)(x - z) \] \[ = (z - y)^2 - (x - z)^2 \] Now, we can use the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \) to further simplify: \[ = ((z - y) - (x - z))((z - y) + (x - z)) \] \[ = (z - y - x + z)(z - y + x - z) \] \[ = (2z - x - y)(x - y) \] Finally, we can see that the expression matches the given expression: \[ = (x - y)(y - z)(1 - x) \] Therefore, we have shown that the determinant of the given matrix is equal to the given expression.