If $x, y, z$ distinct common roots of $z^6 - 1 = 0$ and $z^3 - 1 = 0$ then $$\begin{vmatrix} x - y - z & 2x & 2x \\ 2y & y - x - z & 2y \\ 2z & 2z & z - x - y \end{vmatrix}$$ is equal to _____.
Step-by-Step Solution
Key Concept: The common roots of z^6 - 1 = 0 and z^3 - 1 = 0 are roots satisfying gcd(6,3) = 3, yielding only z = 1 (since z^3 = 1 requires z³ - 1 = 0). When x, y, z are distinct common roots, the constraint x + y + z = 0 or a symmetric property emerges; applying row operation R₁ → R₁ + R₂ + R₃ creates a row of zeros, yielding determinant = 0.
Given $z^6 - 1 = 0$ and $z^{21} - 1 = 0$, the common roots are the 6th roots of unity: $1, \omega, \omega^2$ where $\omega = e^{2\pi i/3}$. From $R_1 = R_1 + R_2 + R_3$ and the determinant equation $\begin{vmatrix} x+y+z & x+y+z & x+y+z \\ 2y & y-x-z & 2y \\ 2z & 2z & z-x-y \end{vmatrix} = 0$, this simplifies to a constraint on $x, y, z$.
Correct Answer: 0