I learned electrodynamics. According to the vector potential determination, $ \mathbf B = [\nabla \times \mathbf A ], $ Coulomb gauge, $ \nabla \mathbf A = 0, $ and one of Maxwell's equations, $ [\nabla \times \mathbf B ] = \frac{1}{c}4\pi \mathbf j, $
I can assume, that
$ [\nabla \times \mathbf B ] = \nabla (\nabla \mathbf A) - \Delta \mathbf A = -\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j. $ How to prove that the one of the solutions of this equation is solution like newtonian potential, $ \mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|}? $