Prove $(\vec A \times \vec B) \cdot (\vec C \times \vec D) = (\vec A \cdot \vec C)(\vec B \cdot \vec D) - (\vec A \cdot \vec D)(\vec C \cdot \vec B)$ Prove that $(\vec A \times \vec B) \cdot (\vec C \times \vec D) = (\vec A \cdot \vec C)(\vec B \cdot \vec D) - (\vec A \cdot \vec D)(\vec C \cdot \vec B)$.
The problem asks to prove this only using the properties:
$
\text{(i)}\space (\vec a \times \vec b) \times \vec c = (\vec a \cdot \vec c)\vec b - (\vec b \cdot \vec c)\vec a \\
\text{(ii)}\space \vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c \\
\text{(iii)}\space \vec u \cdot (\vec v \times \vec w) = \vec v \cdot (\vec w \times \vec u) = \vec w \cdot (\vec u \times \vec v) = -\vec u \cdot (\vec w \times \vec v) = -\vec w \cdot (\vec v \times \vec u) = -\vec v \cdot (\vec u \times \vec w)$
I've tried manipulating the left hand side in all the ways I could think of, and I can't seem to reach the right hand side.
Can someone please point me in the right direction?

Let us omit these horrible vector arrows. Then we have
$$
\begin{align}
( A \times  B) \cdot ( C \times  D) &\overset{(iii)}{=} D\cdot ((A\times B)\times C) \overset{(i)}{=} 
D\cdot( (A\cdot C)B-(B\cdot C)A)\\
&=( A \cdot  C)( B \cdot  D) - ( A \cdot  D)( C \cdot  B),
\end{align}
$$
where in the last equality we have used the symmetry of the scalar product $A\cdot B=B\cdot A$.
