= Solution
Use the <conjugacy class> order of <1/b/solution>. The defining <representation> has <character>
$$
\boxed{\chi_\rho=(2,-2,0,0,0).}
$$
Its <character inner product> with itself is $(4+4)/8=1$, proving that it is an <irreducible representation>. There is also a direct <invariant subspace> argument: $B$ has two distinct coordinate <eigenspaces>, and $A$ exchanges them, so no line is invariant under both.
Let $L_{st}$ denote the <one-dimensional representation> with $L_{st}(A)=s$ and $L_{st}(B)=t$. Then $L_{st}(-I)=1$ and $L_{st}(C)=st$. The complete <character table> is
$$
\boxed{\begin{array}{c|rrrrr}
&I&-I&\pm A&\pm B&\pm C\\\hline
\text{class size}&1&1&2&2&2\\\hline
L_{++}&1&1&1&1&1\\
L_{+-}&1&1&1&-1&-1\\
L_{-+}&1&1&-1&1&-1\\
L_{--}&1&1&-1&-1&1\\
\rho&2&-2&0&0&0
\end{array}}
$$
The four scalar rows obey <character orthogonality>, and the <dimensions> exhaust the <group order> as established in <1/b/solution>.
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