\[ \mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \nabla^2 f \]
where \[g\]
is an arbitrary function.Proof
\[\mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g) + \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \nabla^2 f\]
since
\[\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g)=0 \]
if \[g\]
is twice differentiable.