How can Ishow that $(\vec a \times \vec b) \cdot (\vec a \times \vec b)=|\vec a|^2|\vec b|^2-(\vec a \cdot \vec b)^2$ using index notation? I'm trying to use index notation (i.e. Einstein summation notation) in order to show that  $(\vec a \times \vec b) \cdot (\vec a \times \vec b)=|\vec a|^2|\vec b|^2-(\vec a \cdot \vec b)^2$.
Here's what I've done so far, but I'm stuck at a dead end:
$$\begin{align}(\vec a \times \vec b) \cdot (\vec a \times \vec b)&=(\varepsilon_{ijk}a_jb_k) (\varepsilon_{ijk}a_jb_k) \\ \ \\&=\varepsilon_{ijk}\varepsilon_{ijk}a_ja_jb_kb_k \\ \ \\ &=[\delta_{jj}\delta_{kk}-\delta_{jk}\delta_{kj}] a_ja_jb_kb_k \\ \ \\ &=[3\cdot 3-\underbrace{\delta_{kk}}_{3} \ ]a_ja_jb_kb_k \\ \ \\&=6a_ja_jb_kb_k\end{align}$$ but where do I go from here?
Thanks
 A: Per my comment (which had a typo and should have said $b_jb_m$)
\begin{align}
\varepsilon_{ijk}a_ib_j\hat{e}_k\cdot\varepsilon_{lmn}a_lb_m\hat{e}_n &=
\varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m(\hat{e}_k\cdot\hat{e}_n)\\
&= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m\delta_{kn}\\
&= \varepsilon_{ijk}\varepsilon_{lmk}a_ib_ja_lb_m\\
&= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})a_ib_ja_lb_m
\end{align}
When $i=l$ and $j=m$, we have $a_i^2b_j^2$.  What happens when $i=m$ and $j=l$?

One of the problems you faced by indexing the Levi-Civita symbol both as $\varepsilon_{ijk}$ is when you take the dot product.  Upon taking the dot product, you have 
$$
\hat{e}_k\cdot\hat{e}_k = 1
$$ 
which leads $\varepsilon_{ijk}\varepsilon_{ijk} = 6$.
If you would have had $\varepsilon_{ijk}\varepsilon_{ijn}$, it would have let to same problem since $\hat{e}_k\cdot\hat{e}_n = \delta_{kn}$
A: To clarify my remark in comments: Using any summation index more than twice renders the summation convention is insensible. This becomes obvious if you write in terms of $\Sigma's$: $$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)^2\neq \sum_{j,k=1}^3(\epsilon_{ijk}a_j b_k)^2.$$ What is correct is $$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)\left(\sum_{l,m=1}^3\epsilon_{ilm}a_l b_m\right)= \sum_{jklm}^3 \epsilon_{ijk}a_j b_k\epsilon_{ilm}a_l b_m.$$
So a second pair of indices is absolutely crucial in carrying out the algebra correctly.
