= Solution
Use the index convention $[e_a,e_c]=C_a{}^b{}_c e_b$. For the left- and right-invariant coframes, the <Maurer-Cartan equations> are
$$
\boxed{d\lambda^b=-\frac12C_a{}^b{}_c\lambda^a\wedge\lambda^c,\qquad
d\rho^b=+\frac12C_a{}^b{}_c\rho^a\wedge\rho^c.}
$$
Their dual invariant vector fields satisfy
$$
\boxed{[L_a,L_c]=C_a{}^b{}_cL_b,\qquad
[R_a,R_c]=-C_a{}^b{}_cR_b,\qquad[L_a,R_c]=0.}
$$
For example, evaluating $d\lambda^b$ on two dual fields gives $-\lambda^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.
For a <semisimple Lie algebra>, the <Killing form> $B(X,Y)=\operatorname{Tr}(\operatorname{ad}X\operatorname{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\operatorname{ad}[Z,X]=[\operatorname{ad}Z,\operatorname{ad}X]$ and cyclicity of trace gives
$$
B([Z,X],Y)+B(X,[Z,Y])=0.
$$
Group conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a <bi-invariant pseudo-Riemannian metric>.
Lower the indicated structure index by $C_{abc}=B_{be}C_a{}^e{}_c$. Then
$$
C_{abc}=B(e_b,[e_a,e_c])=-B(e_a,[e_b,e_c]).
$$
This changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence \b[$C_{abc}=C_{[abc]}$].
Let $\eta(X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a <left-invariant differential form>. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the <exterior derivative> on four left-invariant fields. Its six bracket terms combine by invariance into
$$
\begin{aligned}
d\eta(W,X,Y,Z)
&=-2\bigl[B([W,X],[Y,Z])-B([W,Y],[X,Z])+B([W,Z],[X,Y])\bigr]\\
&=-2B\bigl(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\bigr)=0.
\end{aligned}
$$
The final equality is the <Jacobi identity>. Therefore the <Cartan three-form>
$$
\boxed{\frac1{3!}C_{abc}\lambda^a\wedge\lambda^b\wedge\lambda^c}
$$
is both \b[bi-invariant and closed].
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