\alpha_2=(x+y+z)(x+y-z)+ \tau (L)\end{aligned}$$ are also factors of $g$. Since the coefficients of $x^2$ in the polynomials $\alpha_1$ and $\alpha_2$ are the same (and equal 1), and since they are not equal, it follows that $\alpha_1$ and $\alpha_2$ are not associates. By considering the coefficients of $y^2$ and $z^2$, we see that no two of the polynomials $\alpha$, $\alpha_1$ and $\alpha_2$ are associates. Therefore $\alpha \alpha_1 \alpha_2$ divides $g$, a contradiction since $\deg g = 4 < 6$.

Therefore $g$ cannot be reducible.

$$\begin{aligned}
G&=&(x^2+y^2+z^2)^2-3(x^4+y^4+z^4)\\
&=&(-2)(x^2+\omega y^2+ \omega^2 z^2)(x^2+\omega^2 y^2+ \omega z^2),\end{aligned}$$ where $\omega \in k$ is a primitive third root of 1, and where the quadratics on the right hand side are irreducible. Therefore $A$ and $B$ are the polynomials $$u(x^2+\omega y^2+\omega^2 z^2) \mbox{~and~} v(x^2+\omega^2 y^2+\omega z^2),$$ where $$u, v \in k,~uv=-2.$$ Hence $\alpha$ and $\beta$ are the polynomials $$u(x^2+\omega y^2+\omega^2 z^2 +L) \mbox{~and~} v(x^2+\omega^2 y^2+\omega z^2 +K),$$ where $L$ and $K$ are linear polynomials. Also, $$\begin{aligned}
g&=&(a^2+x^2+y^2+z^2)^2-3(a^4+x^4+y^4+z^4) \nonumber\\
&=&(-2)(x^2+\omega y^2+ \omega^2 z^2+L)(x^2+\omega^2 y^2+ \omega z^2+K). \label{Kh1}\end{aligned}$$ Let $L_0$ and $K_0$ be the constant terms of $L$ and $K$, respectively. Plugging $x=y=z=0$ in (\[Kh1\]), we obtain $-2 L_0 K_0 =  -2 a^4$, and hence $$\begin{aligned}
 L_0 K_0 &\ne& 0. \label{Kh2}\end{aligned}$$

