The eigen value problem that must be solved to determine the fcc phonon dispersion relation is, (show derivation)

\begin{equation} \left[ \begin{array}{c} 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \end{array} \right] \left[ \begin{array}{b} u_{\vec{k}}^x \\ u_{\vec{k}}^y \\ u_{\vec{k}}^z \end{array} \right]=\frac{M\omega^2}{C} \left[ \begin{array}{b} u_{\vec{k}}^x \\ u_{\vec{k}}^y \\ u_{\vec{k}}^z \end{array} \right] \end{equation}

The eigen values $\lambda = \frac{M\omega^2}{C}$ can be found by setting the determinant of the matrix to zero.

\begin{equation} \left| \begin{array}{c} 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) -\lambda & - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) -\lambda & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) -\lambda \end{array} \right|=0 \end{equation}

The matrix is symmetric with six independent elements. It has the form,

\begin{equation} \left| \begin{array}{c} m_{11} -\lambda & m_{12} & m_{13} \\ m_{12} & m_{22} -\lambda & m_{23} \\ m_{13} & m_{23} & m_{33} -\lambda \end{array} \right|=0 \end{equation}

This can be written as a cubic equation in $\lambda$.

$$-\lambda^3 +(m_{11}+m_{22}+m_{33})\lambda^2+(m_{12}^2+m_{13}^2+m_{23}^2 - m_{11}m_{22} - m_{11}m_{33} - m_{22}m_{33})\lambda + m_{11}m_{22}m_{33} +2m_{12}m_{13}m_{23} - m_{12}^2m_{33} - m_{13}^2m_{22}-m_{23}^2m_{11} =0$$

Cubic equations can be solved using Cardano's formula. The standard form of a cubic equation is,

$$a\lambda^3 + b\lambda^2 + c\lambda + d =0$$

Code that will calculate the coefficients $a,b,c,d$ and the matrix elements from the $\vec{k}$-vector is,

$k_xa=$  $k_ya=$  $k_za=$  

$\sqrt{\frac{M}{C}}\omega = $

	$\sqrt{M/C}\omega_1$	$\sqrt{M/C}\omega_2$	$\sqrt{M/C}\omega_3$
$\Gamma: 0,0,0$	0	0	0
$X: \frac{2\pi}{a},0,0$	$2$	$2$	$2\sqrt{2}$
$L: \frac{\pi}{a},\frac{\pi}{a},\frac{\pi}{a}$	$\sqrt{2}$	$\sqrt{2}$	$2\sqrt{2}$
$W: \frac{2\pi}{a},\frac{\pi}{a},0$	$2$	$\sqrt{6}$	$\sqrt{6}$
$U: \frac{2\pi}{a},\frac{\pi}{a},\frac{\pi}{a}$	$\sqrt{2}$	$2$	$\sqrt{6}$
$K: \frac{3\pi}{2a},\frac{3\pi}{2a},0$	$1.8477$	$2.3268$	$2.6131$