To establish the required result, begin by considering the square of the left-hand side.

To go from the second to the third equality, note that the terms in red have been both added and subtracted. The resulting grouping allows the positive terms to be recognized as the product of the squares of the norms of A and B; and the negative terms, as the square of the dot product of A and B.

The essential step of adding and subtracting the "right" expression in the calculations in the Mathematical Solution was first discovered by working backwards in Maple. The "magic" by which this discovery was made is articulated below.

If the factors in the first form of the right-hand side are multiplied out, the result is$$

which is nothing more than ${∥\mathbf{A}∥}^2$ ${∥\mathbf{B}∥}^2$ − ${\left(\mathbf{A}·\mathbf{B}\right)}^2$. This is what the expression $q$ must become if ${∥\mathbf{A}\times \mathbf{B}∥}^2$ is to be recognized as the square of $∥\mathbf{A}∥$ $∥\mathbf{B}∥$ $\sin \left(\mathrm{\θ}\right)$. However, $q$ does not immediately factor into this form, and the reason why not has to be determined. To this end, split $q$ into its positive and negative terms. This is done with an application of the selectremove command that returns terms that "have" a given subexpression, and also returns the terms that don't. (The has command returns true or false, either the subexpression is there or it isn't.)

The expression $q$ has been split into $u$, its negative terms, and $v$, its positive terms. The positive terms should factor to ${∥\mathbf{A}∥}^2$ ${∥\mathbf{B}∥}^2$ but they don't. The negative terms should factor to ${\left(\mathbf{A}·\mathbf{B}\right)}^2$, but they don't either. Why not?

Consider the difference between ${∥\mathbf{A}∥}^2$ ${∥\mathbf{B}∥}^2$ and $v$:

The three terms in $Q$ must be added to $v$ for the positive terms in $q$ to factor to ${∥\mathbf{A}∥}^2$ ${∥\mathbf{B}∥}^2$. Indeed, this can be further verified by factoring the sum of $v$ and $Q$:

The three terms in $Q$ must be subtracted from $u$ for the negative terms in $q$ to factor to $-{\left(\mathbf{A}·\mathbf{B}\right)}^2$. Indeed, this can be further verified by factoring the difference of $u$ and $Q$:

 (Note that the three terms in $Q$ are precisely the three red terms added and subtracted in the Mathematical Solution!)

Consequently, the calculations in the Mathematical Solution can be implemented, and the desired result obtained. The algebraic manipulations needed to discover how to proceed are the forte of Maple.