Cubes as the sum of odd integers It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!

There are identitity's given below:
For even cube:
$p^3=(p^2-p+1)+\cdots+(p^2-5)+(p^2-3)+(p^2-1)+(p^2+1)+(p^2+3)+(p^2+5)+\cdots+(p^2+p-1)$
For odd cube:
$q^3=(q^2-q+1)+\cdots+(q^2-6)+(q^2-4)+(q^2-2)+(q)^2+(q^2+2)+(q^2+4)+(q^2+6)+\cdots+(q^2+p-1)$
For, $p=8$ we get:
$8^3=(57+59+61+63+65+67+69+71)$
For, $q=9$ we get:
$9^3=(73+75+77+79+81+83+85+87+89)$
