= Solution
For a finite-dimensional <Lie algebra>, define its <Killing form> by
$$
\boxed{B(x,y)=\operatorname{tr}(\operatorname{ad}x\operatorname{ad}y),\qquad \operatorname{ad}x(z)=[x,z].}
$$
The trace identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ proves symmetry, and linearity of trace proves bilinearity. The <Jacobi identity> gives $\operatorname{ad}[x,y]=[\operatorname{ad}x,\operatorname{ad}y]$. Consequently, by cyclic invariance of trace,
$$
B([x,y],z)=\operatorname{tr}\bigl((\operatorname{ad}x\operatorname{ad}y-\operatorname{ad}y\operatorname{ad}x)\operatorname{ad}z\bigr)
=\operatorname{tr}\bigl(\operatorname{ad}x[\operatorname{ad}y,\operatorname{ad}z]\bigr)=B(x,[y,z]).
$$
This is the invariance identity. Equivalently \b[$B([h,x],y)+B(x,[h,y])=0$] for all three elements.
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