Help with vector cross product identity In EM Physics we were given the problem to show that
$$\vec a \times (\vec b \times \vec c) = \vec b (\vec a\cdot \vec c) - \vec c (\vec a \cdot \vec b).$$
I know first
$$ \vec a \times (\vec b \times \vec c) = \hat e_i \epsilon_{ijk} a_j (\epsilon_{kmn} b_m c_n )$$
but don't know where to go from here. I don’t want to fully expand. If someone can give me pointer that would be great.
 A: A more down-to-earth approach is to prove this identity in three-dimensional space by writing it down in terms of vector components: $\vec{a} = (a_x, a_y, a_z)$, etc.,  and using the expression for the vector product
$$
\vec{a}\times\vec{b} 
=
\left|
\begin{matrix}
\hat{e}_x & \hat{e}_y & \hat{e}_z \\
a_x & a_y & a_z \\
b_x & b_y & b_z
\end{matrix}
\right|
$$
It may seem a bit tedious, but it is straightforward and foolproof.
Update
In some corners this identity is called Bee-Ay-Cee minus Cee-Ay-Bee, which is a simple mnemonic rule for memorizing it.
A: Use the contraction identities of the Levi-Civita symbol. I strongly encourage you to prove these identities yourself as well, I think you will find it worth the effort in the long run.
A: You can use properties of the  Levi-Civita tensor, specifically
$$\epsilon_{kij}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$$
so that
\begin{align} 
\vec a \times (\vec b \times \vec c) &= \hat e_i \epsilon_{ijk} a_j (\epsilon_{kmn} b_m c_n )&\\
& = \hat e_i ( \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm} )a_j b_m c_n &\\
&= \hat e_i \delta_{im}\delta_{jn} a_j b_m c_n -  \hat e_i \delta_{in}\delta_{jm} a_j b_m c_n &\\
&= \hat e_i a_j b_i c_j - \hat e_i a_j b_j c_i \\
&=\vec b (\vec a\cdot \vec c) - \vec c(\vec a\cdot\vec b)
\end{align}
Any steps in between should be straight forward. You may want to double check these indices are in the correct spot. See the link in the answer provided by Puk.
A: I always have trouble with this identity, so here's a fun way to derive it in three-dimensions. It may be argued that this method is a little convoluted, but I find it much easier to remember than the Levi-Civita contraction formula, and much less tedious than working out the components! Let's call the vector $\mathbf{a \times (b \times c) = d}$, and see what we can say about $\mathbf{d}$, using our intuition.
Now, $\mathbf{d}$ must be perpendicular to $\mathbf{a}$ by the definition of the cross-product. Furthermore, $\mathbf{d}$ must also be perpendicular to the vector $\mathbf{(b\times c)}$. From these two facts, you should be able to see that $\mathbf{d}$ must lie in the plane formed by the vectors $\mathbf{b}$ and $\mathbf{c}$! (If you're not convinced, try it out: the first cross-product between $\mathbf{b}$ and $\mathbf{c}$ takes you out of the $\mathbf{bc}-$plane, and the second cross product (with $\mathbf{a}$) has to bring you back onto it, because we're in three dimensions!)
As a result, since $\mathbf{d}$ lies in the plane of the vectors $\mathbf{b}$ and $\mathbf{c}$, it can therefore be written as a linear combination of them: $$\mathbf{d} = \alpha \, \mathbf{b} + \beta \, \mathbf{c},$$ where $\alpha$ and $\beta$ are scalars.
We now use the fact that $\mathbf{d}$ must be linear in $\mathbf{a}$,$\mathbf{b}$, and $\mathbf{c}$, and therefore all the terms on the right hand side must have only one power of each of these vectors respectively. Thus, $\alpha$ must be proportional to $(\mathbf{a\cdot c})$, since it has to be a scalar constructed from $\mathbf{a}$ and $\mathbf{c}$, and similarly $\beta$ must be proportional to $(\mathbf{a\cdot b})$. Thus, $$\mathbf{d} = A\, (\mathbf{a\cdot c}) \mathbf{b} + B\, (\mathbf{a\cdot b}) \mathbf{c},$$ where $A$ and $B$ are two absolute constants (dimensionless numbers) that are independent of the vectors.
Using the fact that $\mathbf{d}$ changes sign if $\mathbf{b}$ and $\mathbf{c}$ are interchanged, you should trivially be able to show that $A=-B$, and so $$\mathbf{d} = A\, \Big( (\mathbf{a\cdot c}) \mathbf{b} - (\mathbf{a\cdot b}) \mathbf{c}\Big).$$ All that's left to do now is to determine $A$, which is easily done by taking a special case (since the above equation is valid for all vectors), so we could set $\mathbf{a} = \hat{\mathbf{x}}, \mathbf{b} = \hat{\mathbf{x}}, \mathbf{c} = \hat{\mathbf{y}},$ for example, and we'd see in this case that $A= 1$, and so $$\mathbf{a \times (b\times c)} = (\mathbf{a\cdot c}) \mathbf{b} - (\mathbf{a\cdot b}) \mathbf{c}$$
A: Rotation of the axes does not affect the cross and dot products. Since the equation is obvious if  $\mathbf{b}  \times  \mathbf{c}$ is zero, we can assume that $\mathbf{b}  \times  \mathbf{c}$ is a nonzero multiple of $\mathbf{i}.$   Now, we can  assume that $\mathbf{b}=(0,r,0)$ and that $\mathbf{c}=(0,s,t),$ with both $r$ and $t$ nonzero. Thus,
$$\mathbf{a} \times (\mathbf{b}  \times  \mathbf{c})=rt(0,a_3,-a_2).$$ As observed in another answer, $\mathbf{a} \times (\mathbf{b}  \times  \mathbf{c})$ is linear combination of $\mathbf{b}$  and $\mathbf{c}.$ Thus, $\mathbf{a} \times (\mathbf{b}  \times  \mathbf{c})=$ $$rt(0,a_3,-a_2)=k_1\mathbf{b} +k_2\mathbf{c} =k_1(0,r,0)+k_2(0,s,t).$$ Clearly, $$k_2=-ra_2=-\mathbf{a}\cdot \mathbf{b}.$$ From  $rta_3=k_1r+k_2t,$ we conclude that
$$k_1=sa_2+ta_3=\mathbf{a}\cdot \mathbf{c}.$$ Observe that the  cross-product calculations above were extremely easy.
A: You want to know when $\epsilon_{ijk} \epsilon_{kmn}$ is different from zero.  A basic property is that $\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} = -\epsilon_{ikj}$ (cyclical and anti-cyclical permutations of the indices).  So to make reasoning simpler, let's rewrite your product as $\epsilon_{kij} \epsilon_{kmn}$.  Note that this can only be non-zero if all indices $1,2,3$ appear exactly once in each term, and since $k$ appears in both this means that in all non-zero terms either $i=m$ and $j=n$ or $i=n$ and $j=m$.  In the first case both $\epsilon$s have the same indices and thus evaluate to the same value giving a product equal $+1$, in the second case they evaluate to the opposite value, and thus their product is $+1\times -1=-1$.  This can be summed up in the identity given in the accepted answer:$$ \epsilon_{ijk}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}. $$
The remainder is identifying the products that remain after simplifying via $a_i \equiv \delta_{ij} a_j$.
