perfect squares....

$\quad { n }^{ 3 }+{ 2n }^{ 2 }\\ ={ n }^{ 2 }\left( n+2 \right)$

Now $n^2$ is a square number. So in order to make ${ n }^{ 3 }+{ 2n }^{ 2 }$ a perfect square , (n+2) must also be a perfect square. So we can conclude that n is 2 less than a perfect square . Now $37^2$ is the greatest square less than 1380. So, there are 37 different values for n. But when we take 1 as a perfect square, we get value of n as 1-2 = -1 . Hence we cannot take 1 in our case . So total values of n which are greater than 0 are 36 .