Simplify $(A\cdot B)^2 - [(A\times B)\times B)] \cdot A$

Question

Simplify $(A\cdot B)^2 - [(A\times B)\times B)] \cdot A$

My attempt:

$(A \cdot B)^2 = A^2 B^2 cos^2(\theta)$

and

$(A\times B) \times B = B (A \cdot B) - A (B \cdot B)$ from Triple Vector Product Formula

I get

$A^2B^2cos^2(\theta) - [B(A \cdot B) - A(B^2)] \cdot A$

Since $A \cdot B = ABcos(\theta)$

This becomes

$A^2B^2cos^2(\theta) - [B(ABcos(\theta)) - A(B^2)] \cdot A$

I'm a bit lost about how to simplify from here on.

Answer 1

$$\|A\|^2\|B\|^2\cos^2(\theta) - [B(\|A\|\|B\|\cos(\theta)) - A(\|B\|^2)] \cdot A$$ $$= \|A\|^2\|B\|^2\cos^2(\theta) - B \cdot A(\|A\|\|B\|\cos(\theta)) + A \cdot A(\|B\|^2)]=\|A\|^2\|B\|^2$$