in (\[uuu\]).

[*Proof.*]{} If $(n,t)=(3,2)$, then $f$ factors as in (\[uuu\]), and we are done. So we assume that $(n,t)\ne (3,2)$, that $f$ has a linear factor, say $g=a_1x_1+\cdots+a_nx_n$, and we seek a contradiction.

Clearly we may assume that $a_1 \ne 0$. By Lemma \[AAA\], $g$ is not symmetric in the variables $x_2$ and $x_3$. Therefore $a_2 \ne a_3$. If $g_1$ is the polynomial obtained from $g$ by interchanging $x_2$ and $x_3$, then $g_1$ is not an associate of $g$, and it is also a factor of $f$. Therefore $gg_1$ is a $(x_2,x_3)$-symmetric factor of $f$ (of degree 2). This contradicts Lemma \[AAA\]. $\Box$

\[CCC\] Let $k$ be a field with ${\mbox{char~}}k \ne 2$, and let $$\begin{aligned}
f &=& (x_1^2+x_2^2+x_3^2)^2-t(x_1^4+x_2^4+x_3^4) \in k[x_1,x_2,x_3], \end{aligned}$$ where $t \in k$ and $t \ne 0$. If $t=2$, then $f$ factors as in (\[uuu\]). If $t \ne 2$, then $f$ is reducible if and only if $t=3$ and $k$ contains a primitive third root $\omega$ of 1. In this case, $$\begin{aligned}
    f&=&(x_1^2+x_2^2+x_3^2)^2 - 3(x_1^4+x_2^4+x_3^4)    \nonumber \\
    &=& (-2)(x_1^2+\omega x_2^2 + \omega^2 x_3^2)(x_1^2+\omega^2 x_2^2 + \omega x_3^2). \label{vvvv}\end{aligned}$$ Also, the factors in the right hand sides of (\[vvvv\]) are irreducible.

[*Proof.*]{} There is nothing to prove 