 Solution

The variables commute, so $x_i{x}_j=x_j{x}_i$. Thus $V$ has dimension six, with basis $x_1^2,x_2^2,x_3^2,x_1{x}_2,x_1{x}_3,x_2{x}_3$, even though nine coefficients occur in the redundant expression in the source. Its representation is the symmetric square of the three-dimensional permutation representation.

The identity fixes all six monomials. A three-cycle fixes none. For a transposition, say $(12)$, exactly $x_3^2$ and $x_1{x}_2$ are fixed; the remaining monomials occur in exchanged pairs. Therefore ${\chi }_{\tilde{R}}=(6,0,2)$ in the conjugacy class order of solution. The character inner products give multiplicitiesConsequentlyThere is also a concrete decomposition: the span of the three squares is one copy of the permutation representation. The span of the three mixed monomials is another, since $x_2{x}_3,x_1{x}_3,x_1{x}_2$ are permuted exactly like their omitted indices $1,2,3$. Within each, the coefficient-sum-zero vector subspace gives a copy of $D$, and the sum of its three basis monomials gives a trivial representation.

A fixed quadratic polynomial must have equal coefficients on each of the two orbits of monomials. Hence its fixed vector subspace isThe two displayed polynomial invariants are linearly independent and account for both singlets. Equivalently, one may use $x_1^2+x_2^2+x_3^2$ and $(x_1+x_2+x_3{)}^2$ as a basis. No constant term belongs to this homogeneous degree-two vector space.