Show that $ \mathbf u^2 \mathbf v^2 = (\mathbf u \cdot \mathbf v)^2 - (\mathbf u \wedge \mathbf v)^2 $ where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.

You can use
$$
\begin{align}
(u\cdot v)^2 - (u\wedge v)^2
& = \frac{1}{4} \left( (uv+vu)^2 - (uv-vu)^2\right) \\
& = \frac{1}{4} \left(uvuv + uvvu + vuuv + vuvu - uvuv + uvvu + vuuv - vuvu \right) \\
& = \frac{1}{2} \left( uvvu + vuuv\right) \\
& = \frac{1}{2} \left( u^2v^2+u^2v^2\right) \\
& = u^2v^2
\end{align}
$$
The third line to the fourth line is true because
$$uvvu = uv^2u = uuv^2 = u^2v^2$$
where we can re-order terms because $u^2$, $v^2$ are scalars.
