$n^3$

The $n^{th}$ row of $n^{th}$ rhombus contains $n$, $n$ times which sum up to $n^2$. The $(n-1)^{th}$ and $(n+1)^{th}$ row of $n^{th}$ rhombus contains $n-1$ and $n+1$ (a total of $2n$), $n-1$ times which sum up to $2n(n-1)$. Similarly the $(n-2)^{th}$ and $(n+2)^{th}$ row of $n^{th}$ rhombus contains $n-2$ and $n+2$ (a total of $2n$), $n-2$ times which sum up to $2n(n-2)$. Hence the sum of all the no.s in the $n^{th}$ rhombus is : $n^2+2n(n-1)+2n(n-2)+\cdots+2n(n-(n-2))+2n(n-(n-1))$ Which is equivalent to : $n^2+2n[n(n-1)-(1+2+\cdots+(n-1))]$ This implies : $n^2+2n[n(n-1)-(\frac{n(n-1)}{2})]$ This is equivalent to : $n^2+2n[\frac{n(n-1)}{2}]$ Or $n^2+n^2(n-1) = n^3$