jmc

The prime factorization of $432=2^4\cdot 3^3$. It follows that the sum of the divisors is equal to $(1+2+2^2+2^3+2^4)(1+3+3^2+3^3)$, as each factor of $432$ is represented when the product is expanded. We have to subtract $432$ so that we only count the proper divisors, so the answer is $$\begin{array}{rl}(1+2+4+8+16)(1+3+9+27)-432& =(31)(40)-432\\ & =1240-432\\ & =\boxed{808}.\end{array}$$