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Solution

If $x, y, z$ are three cube roots of $27$, then the determinant of the matrix \[ \begin{pmatrix} x & y & z\\[4pt] y & z & x\\[4pt] z & x & y \end{pmatrix} \] is: The cube roots of $27 = 3^3$ are: \[ x = 3,\qquad y = 3\omega,\qquad z = 3\omega^2, \] where $\omega$ is a cube root of unity satisfying \[ \omega^3 = 1,\qquad 1+\omega+\omega^2 = 0. \] For a circulant matrix, the determinant is: \[ (x+y+z)(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z). \] Now compute the first factor: \[ x+y+z = 3(1+\omega+\omega^2) = 3\cdot 0 = 0. \] Therefore, \[ \det = 0. \]