[proofplan] The proof uses only the defining property of a Lie algebra representation and the cyclicity of the trace in finite dimension. First we record bilinearity from the linearity of $\rho$, composition, and trace. Then symmetry follows from $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Finally, invariance follows by replacing $\rho([x,y])$ and $\rho([y,z])$ with commutators and cyclically moving factors inside the trace. [/proofplan] [step:Verify that $B_\rho$ is bilinear] Let $a,b \in k$ and let $x_1,x_2,y \in \mathfrak g$. Since $\rho:\mathfrak g \to \mathfrak{gl}(V)$ is $k$-linear, composition in $\operatorname{End}_k(V)$ is bilinear, and $\operatorname{tr}:\operatorname{End}_k(V)\to k$ is $k$-linear, we have \begin{align*} B_\rho(ax_1+bx_2,y) &= \operatorname{tr}\bigl(\rho(ax_1+bx_2)\rho(y)\bigr) \\ &= \operatorname{tr}\bigl((a\rho(x_1)+b\rho(x_2))\rho(y)\bigr) \\ &= a\operatorname{tr}\bigl(\rho(x_1)\rho(y)\bigr) + b\operatorname{tr}\bigl(\rho(x_2)\rho(y)\bigr) \\ &= aB_\rho(x_1,y)+bB_\rho(x_2,y). \end{align*} The same argument in the second variable, using the bilinearity of composition and the linearity of trace, gives \begin{align*} B_\rho(x,ay_1+by_2)=aB_\rho(x,y_1)+bB_\rho(x,y_2) \end{align*} for all $x,y_1,y_2 \in \mathfrak g$ and $a,b \in k$. Hence $B_\rho$ is bilinear. [/step] [step:Prove symmetry by cyclicity of the trace] Let $A,T \in \operatorname{End}_k(V)$. The finite-dimensional trace satisfies \begin{align*} \operatorname{tr}(AT)=\operatorname{tr}(TA). \end{align*} Applying this with $A=\rho(x)$ and $T=\rho(y)$ gives, for all $x,y \in \mathfrak g$, \begin{align*} B_\rho(x,y) &= \operatorname{tr}\bigl(\rho(x)\rho(y)\bigr) \\ &= \operatorname{tr}\bigl(\rho(y)\rho(x)\bigr) \\ &= B_\rho(y,x). \end{align*} Thus $B_\rho$ is symmetric. [/step] [step:Expand the representation identity on commutators] Because $\rho:\mathfrak g \to \mathfrak{gl}(V)$ is a Lie algebra homomorphism and $\mathfrak{gl}(V)$ has bracket $[A,T]=AT-TA$, we have, for all $x,y,z \in \mathfrak g$, \begin{align*} \rho([x,y]) &= \rho(x)\rho(y)-\rho(y)\rho(x), \\ \rho([y,z]) &= \rho(y)\rho(z)-\rho(z)\rho(y). \end{align*} Therefore \begin{align*} B_\rho([x,y],z) &= \operatorname{tr}\bigl(\rho([x,y])\rho(z)\bigr) \\ &= \operatorname{tr}\bigl((\rho(x)\rho(y)-\rho(y)\rho(x))\rho(z)\bigr) \\ &= \operatorname{tr}\bigl(\rho(x)\rho(y)\rho(z)\bigr) - \operatorname{tr}\bigl(\rho(y)\rho(x)\rho(z)\bigr). \end{align*} [/step] [step:Move the trace cyclically to obtain the invariant identity] Using cyclicity of trace in the form \begin{align*} \operatorname{tr}(ABC)=\operatorname{tr}(BCA) \end{align*} for $A,B,C \in \operatorname{End}_k(V)$, we obtain \begin{align*} \operatorname{tr}\bigl(\rho(y)\rho(x)\rho(z)\bigr) = \operatorname{tr}\bigl(\rho(x)\rho(z)\rho(y)\bigr). \end{align*} Substituting this into the previous expression gives \begin{align*} B_\rho([x,y],z) &= \operatorname{tr}\bigl(\rho(x)\rho(y)\rho(z)\bigr) - \operatorname{tr}\bigl(\rho(x)\rho(z)\rho(y)\bigr) \\ &= \operatorname{tr}\bigl(\rho(x)(\rho(y)\rho(z)-\rho(z)\rho(y))\bigr) \\ &= \operatorname{tr}\bigl(\rho(x)\rho([y,z])\bigr) \\ &= B_\rho(x,[y,z]). \end{align*} Thus $B_\rho$ is invariant. Combining bilinearity, symmetry, and invariance proves the theorem. [/step]