Let $V_1$ be the defining 3-dimensional representation of SU(3) with character $\chi_1$. Likewise, let $V_2$ be the conjugate representation with character $\chi_2 = \overline{\chi_1}$. Then every irreducible representation of SU(3) has a character which is polynomial in $\chi_1$ and $\chi_2$.
What does this have to do with your problem? Well, letting $u=\chi_1$ and $v=\chi_2$, your polynomials are simple linear combinations of the polynomials of the irreducible characters of SU(3). For example the polynomial $v^2-u$ corresponds to the character of the irreducible representation which is the symmetric square of $V_2$, which is the (0,2)-representation. Similarly $v^3-2uv+1$ corresponds to the character of the (0,3)-representation which is the third symmetric power of $V_2$. (Here the $(a,b)$-representation is the representation with highest weight $a\omega_1 + b\omega_2$ with $\omega_1$ and $\omega_2$ the highest weights of $V_1$ and $V_2$ respectively).
Now since the character $\chi_{[a,b]}$ of the $(a,b)$-representation is the complex conjugate of the character of the (b,a)-representation, the polynomial $P_{a,b}(\chi_1,\chi_2)$ expressing the character of the $(a,b)$-representation satisfies
$P_{a,b}(\chi_1,\chi_2) = \overline{P_{a,b}(\chi_2,\chi_1)} = P_{b,a}(\chi_2,\chi_1)$
This implies the polynomials $P_{a,b}$ satisfy your condition (if you want a short argument of this fact, give me a little while to think of something coherent).
But the short of it is, your polynomials correspond to characters of SU(3).