y^2+ \lambda^2 z^2),~~
      h = a (x^2+\lambda^2 y^2+ \lambda z^2).$$ Therefore $$\begin{aligned}
 gh &=& a^2 [(x^4+y^4+z^4) - (x^2y^2+y^2z^2+z^2x^2)].\end{aligned}$$ But $$\begin{aligned}
 f &=& (1-t) (x^4+y^4+z^4) +2(x^2y^2+y^2z^2+z^2x^2).\end{aligned}$$ Therefore it follows from $f = c gh$ that $(1-t)=  ca^2$ and $-a^2c = 2$. Hence $1-t=  -2$ and $t=3$. Therefore $$\begin{aligned}
    f&=&(x^2+y^2+z^2)^2-3(x^4+y^4+z^4)  \\
    &=&(-2)(x^2+\lambda y^2 + \lambda^2 z^2)(x^2+\lambda^2 y^2 + \lambda z^2),\end{aligned}$$ as desired.

The two factors in (\[vvvv\]) are irreducible because $f$ has no linear factors, by Lemma \[BBB\]. This completes the proof. $\Box$

\[CCDD\] Let $k$ be a field with ${\mbox{char~}}k \ne 2$, and let $a$ be a non-zero element of $k$. Let $$\begin{aligned}
g &=& (a^2 + x_1^2+\cdots+x_n^2)^2-t(a^4 +x_1^4+ \cdots +x_n^4) \in k[x_1,\cdots,x_n], \end{aligned}$$ where $t \in k$ and $t \ne 0$, and where $n \ge 3$. Then $g$ is irreducible.

[*Proof.*]{} We start with the case $n=3$, and we rename $x_1$, $x_2$, $x_3$ as $x$, $y$, $z$. Thus we are to prove that the polynomial $$g = (a^2 + x^2 + y^2 + z^2)^2 - t(a^4 + x^4 + y^4 + z^4),~~a\ne 0, ~t \ne 0, ~{\mbox{char~}}k \ne 2,$$ is irreducible. The general case will follow easily as shown later.

Suppose that $g$ is reducible, and that $g = \alpha 