Problem source

From the facts 
\begin{alignat*}{2}
1 & \quad=\quad &  & 0\\
2+3+4 & \quad=\quad &  & 1+8\\
5+6+7+8+9 & \quad=\quad &  & 8+27\\
10+11+12+13+14+15+16 & \quad=\quad &  & 27+64
\end{alignat*}
guess a general law. Prove it. 
Hence, or otherwise, prove that 
\[
1^{3}+2^{3}+3^{3}+\cdots+N^{3}=\tfrac{1}{4}N^{2}(N+1)^{2}
\]
for every positive integer $N$. 
[\textbf{Hint. }\textit{You may assume that} $1+2+3+\cdots+n=\frac{1}{2}n(n+1)$.]

Solution source

\begin{align*}
&& (n^2+1) + (n^2+2) + \cdots + (n+1)^2 &= n^3+(n+1)^3 \\
\Leftrightarrow && \sum_{i=n^2+1}^{(n+1)^2} i &= n^3 + (n+1)^3  \\
&& \sum_{i=n^2+1}^{(n+1)^2} i  &= \sum_{i=1}^{(n+1)^2} i- \sum_{i=1}^{n^2} i \\
&&&= \frac{(n+1)^2((n+1)^2+1)}{2} - \frac{n^2(n^2+1)}{2} \\
&&&= \frac{(n+1)^2(n^2+2n+2) - n^2(n^2+1)}{2} \\
&&&= \frac{2(n+1)^3+n^2(n^2+2n+1) - n^2(n^2+1)}{2}\\
&&&= \frac{2(n+1)^3+2n^3 + n^2(n^2+1) - n^2(n^2+1)}{2}\\
&&&= (n+1)^3+n^3
\end{align*}

\begin{align*}
&& \sum_{i=1}^{N^2} i &=(0^3+1^3)+ (1^3+2^3)+(2^3+3^3) + \cdots + ((N-1)^3+N^3) \\
&&&=  2 \left (1^3+2^3 + 3^3 + \cdots + (N-1)^3 \right) + N^3 \\
\Rightarrow && \sum_{i=1}^N i^3 &= \frac12 \left ( N^3+ \sum_{i=1}^{N^2} i  \right) \\
&&&= \frac12 \left ( N^3 + \frac{N^2(N^2+1)}{2} \right) \\
&&&= \frac{N^2(N^2+1)+2N^3}{4} \\
&&&= \frac{N^2(N^2+2N+1)}{4} \\
&&&= \frac{N^2(N+1)^2}{4} \\
\end{align*}