Find the smallest number abc

$100a+10b+c=a+b^2+c^3 \implies 99a=c(c-1)(c+1)-b(10-b)$

I think, an straight forward approach is to make tables for possible values of $c(c-1)(c+1)$ and $b(10-b)$ and see if the any of their differences is a multiple of $99$ . for $c=5$ and $b=3$ , $c(c-1)(c+1)=120$ and $b(10-b)=21$ , respectively. so, $c(c-1)(c+1)-b(10-b)=99$ , which gives the least possible value for $a$ , which is $1$ .