Special AP

Since $a,b,c,d,e$ form an arithmetic progression we let

$a=c-2x \ , \ b=c-x \ , \ d=c+x \ , \ e=c+2x$

So we have :

$a+b+c+d+e=5c \\ b+c+d=3c$

In order to have $5c$ as perfect cube , $c$ must contain $5^k$ where $k\equiv 2 \pmod{3}$ and to have $3c$ as a perfect square , $c$ must contain a $3^m$ such that $m$ is odd in its prime factorization. But since $5c$ is a perfect cube , $m$ must be divisible by $3$ .

The least such possible $c=3^3 \times 5^2 = 27 \times 25 = \boxed{675}$