Vector's proof question Two vectors $v, w $ are non-parallel and non-zero.
We need to show that
$$v \bullet(w \times (v \times w) = |w|^2 |v|^2 - (w \bullet v)^2 $$
So far, this is what I have come up with for the proof: 
$$v \bullet(w \times (v \times w) = |w|^2 |v|^2 - (w \bullet v)^2  $$ 
Let $b =  (v \times w) $
Then:
$$v \bullet(w \times b) = |w|^2 |v|^2 - (w \bullet v)^2  $$
$$(v \times w) \bullet b = |w|^2 |v|^2 - (w \bullet v)^2  $$ 
Replace $b$ with $ (v \times w) $
$$(v \times w) \bullet (v \times w) = |w|^2 |v|^2 - (w \bullet v)^2  $$
$$(v \times w) \bullet (v \times w) = (w \bullet w)(v \bullet v)- (w \bullet v)^2  $$
$$(v \times w) \bullet (v \times w) = (w \bullet w)(v \bullet v)- (w \bullet v)(w \bullet v)  $$
This is as far as I can get. 
I don't quite understand what I can do at this point as I don't think I can change the cross products into dot products. 
I have looked through all the dot product and cross product laws and can find none that might convert especially the left side to dot products.
Any ideas?                
 A: The problem with your proof is that you are starting with the equality you want to end up with and are carrying the equality down. That is a big no-no. Logically you can do one of two things: either work your way from one side to the other, or manipulate both sides independently until they both look like the same thing.
Putting that aside, let's start with the left side and try to work our way over to the right. For this we need the famous "bac-cab" rule:
$$a\times(b\times c) = b(a\cdot c) - c(a\cdot b)$$
Applying this rule, we get:
$$v \cdot (w \times (v \times w)) = v \cdot (v (w \cdot w) - w (v \cdot w)) = (v \cdot v)(w \cdot w) - (v \cdot w)^2 = |v|^2|w|^2 - (v\cdot w)^2$$
A: Notice that $$\vec A \times (\vec B \times \vec C)=(\vec A. \vec C)\vec B -(\vec A. \vec B) \vec C.$$ Then
$$\vec v . (\vec w \times (\vec v \times \vec w))=\vec v .[\vec w . \vec w) \vec v-(\vec w.\vec v) \vec w] = |\vec w|^2 |\vec v|^2- (\vec w .\vec v)^2.$$ Hece proved.
Note that your starting statement that the box product $\vec v. (\vec w \times \vec b)=|w|^2 |v|^2- (\vec w . \vec v)^2$ is not correct
