The discriminant of variables $x_1, \dots, x_n$,
$$d=d(x_1, \dots, x_n)=\prod_{i<j}(x_i-x_j)$$
is a classical example of a polynomial whose square $d^2$ is symmetric in its entries. This turns out to be very useful for understanding what do some field extensions look like. So the question is:
Does there exist a polynomial $p(x_1, \dots,x_n)$ which is not symmetric but $p^3$ is? More generally, for some $k>1$, is there a polynomial such that $p, p^2, \dots, p^{k-1}$ are not symmetric and $p^k$ is?
Best Answer
Nice question! The answer is no, there are no such polynomials.
Recall that if $F$ is a field, $x_1, \dots x_n$ are indeterminates and $e_1, \dots e_n$ are the elementary symmetric polynomials in the $x_i$, then $F(x_1, \dots x_n)$ is a Galois extension of $F(e_1, \dots e_n)$ with Galois group $S_n$ acting by permutation on the $x_i$ (in fact it is the splitting field of the polynomial $\prod (t - x_i)$ whose coefficients are the $e_i$). Below we'll need $\text{char}(F) \neq 2$ because we'll be dealing with a lot of minus signs.
In this language the existence of the discriminant can be explained as follows. Suppose there is a polynomial $p \in F(x_1, \dots x_n)$ which is not symmetric but such that $p^2$ is symmetric. Then $p$ would be a root of the quadratic polynomial $t^2 - p^2$, which is an irreducible polynomial over $F(e_1, \dots e_n)$ splitting over $F(x_1, \dots x_n)$. This means that the Galois group $S_n$ acts transitively on its roots, so for any $\sigma \in S_n$ we would have that either $\sigma p = p$ or $\sigma p = -p$, and the latter case must occur at least once. This produces a nonzero homomorphism $S_n \to \{ \pm 1 \}$, and it's a classic result that there is exactly one such homomorphism, the sign homomorphism. So $p$ must be a polynomial which satisfies $\sigma p = \text{sgn}(\sigma) p$, generating the unique quadratic extension of $F(e_1, \dots e_n)$ contained in $F(x_1, \dots x_n)$, which by the fundamental theorem of Galois theory must be the fixed field $F(x_1, \dots x_n)^{A_n}$ of the unique subgroup of index $2$ in $S_n$, namely the alternating group $A_n$, the kernel of the sign homomorphism.
(More explicitly, because $S_n$ is generated by transpositions and every transposition is conjugate, the above condition is equivalent to the condition that $p$ is sent to $-p$ by any transposition of two of the variables $x_i$. This actually implies that $p$ must vanish if any two of the variables $x_i$ are set equal, so as a polynomial it must be divisible by $x_i - x_j$ for all $i \neq j$!)
Now suppose $p$ is a polynomial such that $p^k$ is symmetric, $k \ge 3$. Arguing as above, $p$ is a root of the polynomial $t^k - p^k$, and so again $S_n$ acts on the roots of this polynomial (but not necessarily transitively). So if $\sigma \in S_n$ is a permutation then $\sigma p$ is another root of this polynomial, meaning $(\sigma p)^k = p^k$, which gives that $\frac{\sigma p}{p}$ is a $k^{th}$ root of unity. This produces a homomorphism
$$S_n \ni \sigma \mapsto \frac{\sigma p}{p} \in \mu_k(F) = \{ \zeta \in F : \zeta^k = 1 \}$$
from $S_n$ to the group of $k^{th}$ roots of unity in $F$. This group is abelian, so this homomorphism factors through the abelianization of $S_n$, but this abelianization is just $C_2$ (and the abelianization map $S_n \to C_2$ is the sign homomorphism). This gives that we must in fact have $\sigma p = \pm p$ for all $\sigma \in S_n$ exactly as above, meaning that $(t - p)(t + p) = t^2 - p^2$ must already have coefficients in $F(e_1, \dots e_n)$, so $p^2$ must already be symmetric.