3$. Then $f$ has a factor that is symmetric in two of the variables if and only if $n=3$ and $t=2$. In this case, $$\begin{aligned}
f &=& (x_1^2+x_2^2+x_3^2)^2-2(x_1^4+ x_2^4+ x_3^4)\nonumber \\&=&
(x_1+x_2+x_3)(-x_1+x_2+x_3)(x_1-x_2+x_3)(x_1+x_2-x_3). \label{uuu}
 \end{aligned}$$

[*Proof.*]{} We shall freely use the fact that factors of a homogeneous polynomial are homogeneous; see (\[hhh\]).

Let $$\begin{aligned}
&&y = x_{n-1}, ~ z = x_n,~ u = y+z,~v=yz, \label{yzuv}\\
&&S_2=x_1^2+\cdots+x_{n-2}^2,~S_4=x_1^4+\cdots+x_{n-2}^4,\label{S2S4}\\
&&R=k[x_1,\cdots,x_n] = k[x_1,\cdots,x_{n-2},y,z],~
 R_0=k[x_1,\cdots,x_{n-2},u,v]. \label{RR}\end{aligned}$$ Clearly, $R$ and $R_0$ are polynomial rings over $k$ in the $n$ respective variables, and $R_0$ is the subring of $R$ consisting of the $(y,z)$-symmetric polynomials, i.e., polynomials that are symmetric in the variables $y$ and $z$; see (\[symsym\]).

Using the identities $$\begin{aligned}
y^2+z^2&=&u^2-2v,\\
y^4+z^4&=&(y^2+z^2)^2-2v^2~=~(u^2-2v)^2-2v^2,
$$ and the definition $$f=(S_2+y^2+z^2)^2-t(S_4+y^4+z^4),$$ we obtain (after few lines of computations, or by Maple) $$\begin{aligned}
 \label{fSuv}
f&=&(4-2t)v^2 -4 ((1-t)u^2+S_2)v +S_2^2 + (1-t) u^4  + 2S_2 u^2 -t S_4.\end{aligned}$$ Letting $F \in R_0$ denote the right hand side of (\[fSuv\]), i.e., $$\begin{aligned}
F&=&  (4-2t)v^2 -4 ((1-t)u^2+S_2)v +S_2^2 + (1-t) u^4  + 2S_2 u^2 -t S_4, \label{F}\end{aligned}$$ we see that $f$ has a $(y,z)$-symmetric factor if and only if $F$ is reducible in $R_0$. From now on, we will be working in $R_0$.

Suppose that $F$ 