Cross product: does Jacobi identity imply Lagrange identity? A cross product can be defined as a bilinear operation on a  real vector space with inner product that has the
property of orthogonality: $ \mathbf u \cdot(\mathbf u \times \mathbf v)=\mathbf v \cdot(\mathbf u \times \mathbf v)=0$
and the Pythagorean property $ |\mathbf u \times \mathbf v|^2=u^2v^2-(\mathbf u \cdot \mathbf v)^2$
Starting from such definition we can prove that a non trivial cross product can exists only on $\mathbb R^3$ (where we have two possible cross products with opposite orientation) and on $\mathbb R^7$ (where we can have 480 different cross products. See:  https://en.wikipedia.org/wiki/Seven-dimensional_cross_product).
This question is motivated by the fact that  I found (see: https://www.jstor.org/stable/2315620) the statement that in seven dimension space the cross product does not satisfy the Jacobi identity.
I know that the Jacobi identity (J).
$$
 \mathbf u \times(\mathbf v\times \mathbf w)+ \mathbf w \times(\mathbf u\times \mathbf v)+ \mathbf v \times(\mathbf w\times \mathbf u)=0
 $$
can be provided as a consequence of the Lagrange identity (V).
$$\mathbf u \times(\mathbf v\times \mathbf w)=(\mathbf u\cdot\mathbf w)\mathbf v-(\mathbf u\cdot\mathbf v)\mathbf w $$
And the proof of (V)$\Rightarrow$(J) can be done without the use of coordinate, only using te properties that define the cross and dot product (see e.g. Geometric proof for triple vector product Jacobi identity).
I can prove that in the space $\mathbb R^7$, with any one of the 480 possible cross products, the Lagrange identity is not valid. (using an argument that does not use the coordinates, similar to this one: Do the BAC-CAB identity for triple vector product have some intepretation?)
So in $\mathbb R^7$ we cannot use the same argument of $\mathbb R^3$ to proof the Jacobi identity. But, to be sure that the Jacobi identity is not valide in 7-D space we need a proof that, in general, (J)$\Rightarrow$ (V). But I don't see how to find this proof.
 A: I found that the proof of (J) $\Rightarrow$ (V) can be done starting from the identity
$$
\mathbf u \times (\mathbf v\times \mathbf w)-\mathbf w \times (\mathbf u\times \mathbf v)=2(\mathbf u\cdot \mathbf w)\mathbf v
-(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w
$$
( proved in  https://www.jstor.org/stable/2315620 , formula (2))
using this in the first term, the Jacobi identity:
$$
\mathbf u \times(\mathbf v\times \mathbf w)+ \mathbf w \times(\mathbf u\times \mathbf v)+ \mathbf v \times(\mathbf w\times \mathbf u)=0
 $$
becomes:
$$
2 \mathbf w \times(\mathbf u\times \mathbf v)+2(\mathbf u\cdot \mathbf w)\mathbf v
-(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w+ \mathbf v \times(\mathbf w\times \mathbf u)=0
$$
and using the same identity on the last triple product:
$$
2 \mathbf w \times(\mathbf u\times \mathbf v)+2(\mathbf u\cdot \mathbf w)\mathbf v
-(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w+
\mathbf u \times(\mathbf v\times \mathbf w)+2(\mathbf v\cdot \mathbf u)\mathbf w-(\mathbf u\cdot \mathbf w)\mathbf v-(\mathbf v\cdot \mathbf w)\mathbf u=0
$$
adding the similar terms,
$$
2 \mathbf w \times(\mathbf u\times \mathbf v)+\mathbf u \times(\mathbf v\times \mathbf w)+(\mathbf u\cdot \mathbf w)\mathbf v+(\mathbf v\cdot \mathbf u)\mathbf w-2(\mathbf v\cdot \mathbf w)\mathbf u=0
$$
and using the  same identity for the second term we find
$$
3\mathbf w \times(\mathbf u\times \mathbf v)+3(\mathbf u\cdot \mathbf w)\mathbf v-3(\mathbf v\cdot \mathbf w)\mathbf u=0
$$
that is the lagrange identity (V)
$$
\mathbf w \times(\mathbf u\times \mathbf v)=(\mathbf v\cdot \mathbf w)\mathbf u-(\mathbf u\cdot \mathbf w)\mathbf v
$$
And, since it is easy to prove that (V) is true in $\mathbb R^3$ but false in $\mathbb R^7$, this prove also that te Jacobi identity is true in $\mathbb R^3$ and false in $\mathbb R^7$.
