in the case when $t=2$, since this is covered in Lemma \[BBB\]. Thus we assume that $t \ne 2$.

Let $x_1, x_2, x_3$ be renamed as $x, y, z$, and suppose that $f$ is reducible and that $g$ is an irreducible factor of $f$. Since $t \ne 2$, Lemma \[BBB\] implies that $g$ is not linear. Thus $g$ is an irreducible quadratic. Thus $$\begin{aligned}
g&=&ax^2+by^2+cz^2+\alpha yz + \beta zx + \gamma xy,\end{aligned}$$ where $a, b, c, \alpha, \beta, \gamma$ are in $k$. Also, $$\begin{aligned}
f&=& (x^2+y^2+z^2)^2 - t(x^4+y^4+z^4)\\
&=& (1-t)(x^4+y^4+z^4)+2(x^2y^2+y^2z^2+z^2x^2).\end{aligned}$$ Letting $s$ be the permutation $s = (x \mapsto y \mapsto z \mapsto x)$, we see that $f$ is divisible by $g$, $s(g)$, and $s^2 (g)$. Since $\deg (g s(g) s^2(g)) = 6 > \deg f$, and since $g, s(g), s^2(g)$ are irreducible, it follows that two (and hence all) of the polynomials $g$, $s(g)$, and $s^2 (g)$ are associates (i.e., constant multiples of each other). Thus $g = \lambda s(g)$ for some $\lambda \in k$. Since $$\begin{aligned}
s(g)&=&ay^2+bz^2+cx^2+\alpha zx + \beta xy + \gamma yz,\end{aligned}$$ it follows that $$b = \lambda a,~ c = \lambda b,~a = \lambda c,~
   \beta = \lambda \alpha,~\gamma = \lambda \beta,~
   \alpha = \lambda 