Algebraic vector proof of Lagrange's identity $$|v · w|
^2 + |v × w|
^2 = |v|^2 |w|^2 $$
Edit
Despite doing it multiple times it seems I have made a meal of the expansion see Jean-Claude's answer for a great explanation
Using $v = (v_1,v_2,v_3)$ and $w = (w_1, w_2, w_3)$ i have expanded the LHS and gotten
$$(v_2)^2(w_3)^2 + (v_3)^2(w_2)^2 + (v_3)^2(w_1)^2 + (v_1)^2(w_3)^2 + (v_1)^2(w_2)^2
 + (v_2)^2(w_1)^2 +(v_1)^2(w_1)^2 + (v_2)^2(w_2)^2 + (v_3)^2(w_3)^2 -\mathbf{2(v_2 w_3 w_2 v_3 + v_3 w_1 v_1 w_3 + v_1 w_2 v_2 w_1)}$$
Now this is RHS minus the bolded terms and I dont know how to get rid of the bolded terms.

You have
\begin{align*}
|v\cdot w|^2 &= (v_1w_1+v_2w_2+v_3w_3)^2  \\
  &= v_1^2w_1^2 + v_2^2w_2^2 + v_3^2w_3^2 + 2v_1w_1v_2w_2 + 2v_2w_2v_3w_3 + 2v_3w_3v_1w_1  \text{.}  \\
v\times w &= \begin{pmatrix}v_2w_3-v_3w_2 \\ v_3w_1-v_1w_3 \\ v_1w_2-v_2w_1\end{pmatrix}  \text{.}  \\
|v\times w|^2 &= (v_2w_3-v_3w_2)^2+ (v_3w_1-v_1w_3)^2 +(v_1w_2-v_2w_1)^2  \\
  &= v_2^2w_3^2+v_3^2w_2^2-2v_2w_2v_3w_3+v_3^2w_1^2+v_1^2w_3^2-2v_1w_1v_3w_3 \\
  &\qquad {} + v_1^2w_2^2+v_2^2w_1^2-2v_1w_1v_2w_2  \text{.}  \\
|v\cdot w|^2+|v\times w|^2 &= v_1^2w_1^2+v_2^2w_2^2+v_3^2w_3^2+v_2^2w_3^2+v_3^2w_2^2+v_3^2w_1^2+v_1^2w_3^2  \\
  &\qquad {} + v_1^2w_2^2+v_2^2w_1^2  \\
  &= v_1^2(w_1^2+w_2^2+w_3^2)+v_2^2(w_1^2+w_2^2+w_3^2)+v_3^2(w_1^2+w_2^2+w_3^2)  \\
  &= (v_1^2+v_2^2+v_3^2)(w_1^2+w_2^2+w_3^2)  \\
  &= |v|^2|w|^2  \text{.}
\end{align*}
See also Lagrange's identity on Wikipedia, for other approaches.
