of the same degree as $\alpha$. Since $h=AB$, and $h \ne 0$, it follows that $A\ne 0$ and therefore $h$ is reducible, a contradiction.

This completes the proof. $\Box$

\[DDD\] Let $k$ be a field with ${\mbox{char~}}k \ne 2$, and let $$\begin{aligned}
f &=& (x_1^2+\cdots+x_n^2)^2-t(x_1^4+ \cdots +x_n^4) \in k[x_1,\cdots,x_n], \end{aligned}$$ where $t \in k$ and $t \ne 0$, and where $n \ge 3$. Then $f$ is reducible in and only in the following two cases:

-   $n=3$ and $t=2$, in which case $f$ is as given in (\[uuu\]).

-   $n=3$, $t=3$, and $k$ contains a primitive third root $\omega$ of unity, in which case $f$ is as given in (\[vvvv\]).

[*Proof.*]{} The case when $n=3$ was completely treated in Lemmas \[BBB\] and \[CCC\]. So we assume that $n \ge 4$. For any polynomial $F \in k[x_1,\cdots,x_n]$, let $F^* \in k[x_1,x_2,x_3]$ be the polynomial obtained from $F$ by putting $x_4=1$ and $x_j = 0$ for all $j \ge 5$. Then $f^*$ is the case $n=3$ and $a=1$ of Theorem \[CCDD\], and is hence irreducible. To show that $f$ is irreducible, we suppose that $f = gh$, and we show that $g$ or $h$ is a constant.

Since $f^*=g^*h^*$, and since $f^*$ 