a,b,c are vectors. How to prove vector b = c, given a.b = a.c and axb = axc?
I think we need an additional assumption that $\displaystyle \mathbf{a} \ne \mathbf{0}$; otherwise the claim is not true.
So assuming $\displaystyle \mathbf{a} \ne \mathbf{0}$, as a first step let's see if we can show that if $\displaystyle \mathbf{a} \times \mathbf{v} = \mathbf{0}$ and $\displaystyle \mathbf{a} \cdot \mathbf{v} = 0$ then $\displaystyle \mathbf{v} = \mathbf{0}$.
We have
$\displaystyle \mathbf{0} = \mathbf{v} \times (\mathbf{a} \times \mathbf{v}) = \mathbf{a} (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v} (\mathbf{a} \cdot \mathbf{v}) = \mathbf{a} (\mathbf{v} \cdot \mathbf{v}) - \mathbf{v} (0) = \mathbf{a} (\mathbf{v} \cdot \mathbf{v}) $
so
$\displaystyle \mathbf{v} \cdot \mathbf{v} = 0$, hence $\displaystyle \mathbf{v} = \mathbf{0}$.
Finally, to show that if $\displaystyle \mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}$ and $\displaystyle \mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c}$ then $\displaystyle \mathbf{b} = \mathbf{c}$, let $\displaystyle \mathbf{v} = \mathbf{a} - \mathbf{c}$ above.