For cross product we have Lagrange's formula: $$\vec a \times (\vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b).$$
The formula can be verified by calculating in coordinates. I was thinking about a more geometrical approach and I have noticed this. If we denote $\vec v= \vec a \times (\vec b\times \vec c)$, then:
- The vector $\vec v$ is perpendicular to $\vec b\times \vec c$, which means that is is a linear combination1 of $\vec b$ and $\vec c$. So we have $$\vec v= \beta\vec b+ \gamma \vec c$$ for some real numbers $\beta$ and $\gamma$.
- This vector is also perpendicular to $\vec a$, which means that $$\vec a \cdot \vec v = 0 = \beta \vec a \cdot \vec b + \gamma \vec a \cdot \vec c.$$
The above equation suffices to determine the vector $\vec v$ up to a scalar multiple, since we know the ratio between $\beta$ and $\gamma$. In this way we get $$\vec a \times (\vec b \times \vec c) = K\left(\vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)\right) \tag{*}$$ where $K$ is some real number, still to be determined.
Is there some simple trick which helps to determine $K$ in $(*)$?
1To be more precise, this argument only works if $\vec b\times\vec c\ne\vec 0$, i.e., if $\vec b$ and $\vec c$ are linearly independent. But if $\vec b \times \vec c$ both sides in $(*)$ are equal to zero, to the equation $(*)$ holds in this case, too.
Here's an idea:
With the scalar triple product, we have $$ b \cdot v = b \cdot [a \times (b \times c)] = (b \times c) \cdot (b \times a) $$ Now, with the formula at the end of this section (which apparently can be seen as the three dimensional case of the Binet-Cauchy identity), we have $$ (b \times c) \cdot (b \times a) = (b \cdot b)(a \cdot c) - (b \cdot a)(c \cdot b) =\\ (b \cdot b) (a \cdot c) - (b \cdot c)(a \cdot b) $$ On the other hand, we have $$ b \cdot v = K b \cdot \left(\vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)\right) =\\ K[(b \cdot b)(a\cdot c) - (b \cdot c)(a \cdot b)] $$