Applying the permutation $\sigma : x \mapsto y \mapsto x$, we obtain $$\begin{aligned}
g&=&(a^2+x^2+y^2+z^2)^2-3(a^4+x^4+y^4+z^4)\\
&=&(-2)(y^2+\omega x^2+ \omega^2 z^2+ \sigma (L))(y^2+\omega^2 x^2+ \omega z^2+\sigma (K)),\\
&=&(-2)(x^2+\omega^2 y^2+ \omega z^2+ \omega^2 \sigma (L))(x^2+\omega y^2+ \omega^2 z^2+\omega \sigma (K)).\end{aligned}$$ Since $k[x,y,z]$ is a unique factorization domain, it follows that $$\begin{aligned}
\sigma (L) = \omega K,~ \sigma (K) = \omega^2 L. \label{Kh3}\end{aligned}$$ Similarly, if $\tau$ is the transposition $x \mapsto z \mapsto x$, then $$\begin{aligned}
\tau (L) = \omega^2 K,~ \tau (K) = \omega L. \label{Kh4}\end{aligned}$$ Observing that the constant terms of $L$ and $K$ are unchanged under permutations on $x, y, z$, and using (\[Kh3\]) and (\[Kh4\]), we obtain $$\begin{aligned}
L_0 = \omega K_0,~L_0 = \omega^2 K_0.\end{aligned}$$ It follows that $L_0 = K_0 =0$, contradicting (\[Kh2\]).

Thus we have proved the case $n=3$. If $n \ge 3$, we let $h$ be obtained from $g$ by putting $x_j = 0$ for all $j \ge 4$. Since $h$ is irreducible by the case $n=3$, it follows that $g$ is irreducible. In fact, if $g=\alpha \beta$, and if $A$ and $B$ are obtained from $\alpha$ and $\beta$ by plugging $x_j = 0$ for all $j \ge 4$, then either $A$ is zero or $A$ is homogeneous 