Triple Cross Product and Triple Commutator Relationship

So I notice that,

$[A,BC] = B[A,C] - [B,A]C$

$A \times (B \times C) = B(A \cdot C) - (B \cdot A)C$

I can't see any reason for them to be the same other than chance. Is this just pure chance or is there something behind it?

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The first identity is wrong; it should read $[A, BC] = B[A, C] + [A, B] C$, and this just expresses the fact that $[A, -]$ is a derivation (en.wikipedia.org/wiki/Derivation_(abstract_algebra)). Roughly speaking the reason this is true is that $[A, -]$ is the infinitesimal version of an inner automorphism: see qchu.wordpress.com/2011/02/26/… . –  Qiaochu Yuan Feb 7 '12 at 9:12
If there remains a relationship after you take Qiaochu's comment into account, it probably has something to do with the cross product being isomorphic to the commutator in $so(3)$. –  Rahul Feb 7 '12 at 9:18
@Yuan, fixed, thanks. I'll look into it Narain. –  Stuart Feb 7 '12 at 9:26