I'm reading a passage from Geometric Algebra for Physicists, and in section 2.4.1 there's a train of logic I'm not quite understanding. I'll copy it word for word, with italics being my comments/concerns:
To explore [geometric products] further, let's form the product of the vector $\mathbf{a}$ with the bivector $\mathbf{b}\wedge\mathbf{c}$. From the associative and distributive properties of the geometric product we have: $\mathbf{a}(\mathbf{b}\wedge\mathbf{c})=\mathbf{a}\frac{1}{2}(\mathbf{b}\mathbf{c}-\mathbf{c}\mathbf{b})=\frac{1}{2}(\mathbf{a}\mathbf{b}\mathbf{c}-\mathbf{a}\mathbf{c}\mathbf{b})$. (I understand this part) We now use the rearrangement $\mathbf{a}\mathbf{b}=2\mathbf{c}\cdot\mathbf{b}-\mathbf{b}\mathbf{a}$ (I don't get this step) to write $\mathbf{a}(\mathbf{b}\wedge\mathbf{c})=(\mathbf{a}\cdot\mathbf{b})\mathbf{c}-(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-\frac{1}{2}(\mathbf{b}\mathbf{a}\mathbf{c}-\mathbf{c}\mathbf{a}\mathbf{b})=2(\mathbf{a}\cdot\mathbf{b})\mathbf{c}-2(\mathbf{a}\cdot\mathbf{c})\mathbf{b}+\frac{1}{2}(\mathbf{b}\mathbf{c}-\mathbf{c}\mathbf{b})\mathbf{a}$. (I definitely don't understand this one. Where did the 2s come from and the sign change?) Therefore $\mathbf{a}(\mathbf{b}\wedge\mathbf{c})-(\mathbf{b}\wedge\mathbf{c})\mathbf{a}=2(\mathbf{a}\cdot\mathbf{b})\mathbf{c}-2(\mathbf{a}\cdot\mathbf{c})\mathbf{b}$.(It suffices to say that I really don't understand this step because I don't get the prior two.)
So this arose because my book had a misprint in an equation. Once that was pointed out the equations make sense. Thanks to @Peter for the help.