Mathematics

The key insight is recognizing that $f(f(x)) - x = 0$ factors as $(f(x) - x) \cdot q(x) = 0$, where $q(x)$ is a quadratic whose roots are the "new" fixed points $\gamma$. Since $f(x) - x$ has imaginary roots $\alpha, \beta$, they're also roots of $f(f(x)) - x$, making them appear in the determinant's structure. The determinant's value depends on the algebraic relationships between these roots—specifically, that $\alpha + \beta$ and $\alpha\beta$ are real (from Vieta's formulas for the quadratic $f(x) - x$), which constrains the determinant to a specific value independent of the particular coefficients $a, b, c$.

The key insight is recognizing that $f(f(x)) - x = 0$ factors as $(f(x) - x) \cdot q(x) = 0$, where $q(x)$ is a quadratic whose roots are the "new" fixed points $\gamma$. Since $f(x) - x$ has imaginary roots $\alpha, \beta$, they're also roots of $f(f(x)) - x$, making them appear in the determinant's structure. The determinant's value depends on the algebraic relationships between these roots—specifically, that $\alpha + \beta$ and $\alpha\beta$ are real (from Vieta's formulas for the quadratic $f(x) - x$), which constrains the determinant to a specific value independent of the particular coefficients $a, b, c$.