\beta$, where $\alpha$ and $\beta$ are non-constant polynomials in $k[x,y,z]$. Let $G$, $A$, and $B$ be the leading homogeneous components of $g$, $\alpha$, and $\beta$, respectively. Then $$G=(x^2+y^2+z^2)^2-t(x^4+y^4+z^4),$$ and $G = AB$, and $A$ and $B$ are non-constant. Therefore $G$ is reducible. By Lemma \[CCC\], we have the following two cases:

$$\begin{aligned}
G&=&(x^2+y^2+z^2)^2-2(x^4+y^4+z^4)\nonumber \\
&=&(x+y+z)(-x+y+z)(x-y+z)(x+y-z), \label{GUV}\end{aligned}$$ and hence $$\begin{aligned}
g&=&(a^2+x^2+y^2+z^2)^2-2(a^4+x^4+y^4+z^4).\end{aligned}$$

Suppose that one of the factors $\alpha$ or $\beta$, say $\alpha$, is linear. Then we may assume that either $\alpha=x+y+z+c$ or $\alpha=x+y-z+c$, where $c \in k$. In the first case, we plug $(x,y,z)=(x,-x,-c)$ in $g = \alpha \beta$, and in the second case, we plug $(x,y,z)=(x,-x,c)$. In both cases, we obtain $$\begin{aligned}
0&=&(a^2+2x^2+c^2)^2-2(a^4+2x^4+c^4)
\\&=& 4x^2(a^2+c^2) +(-a^4-c^4+2a^2c^2)\\
&=& 4x^2(a^2+c^2)-(a^2-c^2)^2.\end{aligned}$$ Therefore $a^2+c^2=a^2-c^2=0$, and hence $a=c=0$, contradicting the assumption that $a \ne 0$.

If $\alpha$ and $\beta$ are quadratic irreducible, then $A$ and $B$ are quadratic. Also $G=AB$ by (\[fgh\]). By (\[GUV\]), we may assume that $A$ or $B$, say $A$, is $(x+y+z)(-x+y+z)$. Thus $$\begin{aligned}
\alpha&=&(x+y+z)(-x+y+z)+L,\end{aligned}$$ where $L$ is a linear polynomial. By applying the permutation $\sigma : x \mapsto y \mapsto x$ and then $\tau : x \mapsto z \mapsto x$, we see that the two (irreducible) polynomials $$\begin{aligned}
\alpha_1=(x+y+z)(x-y+z)+\sigma (L) \mbox{~~and~~} 