Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity  $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.

I find it interesting but have trouble seeing the proof behind it.  Could anyone could give me a formal proof based on this animation?

Method 1 (By $k^3=k\cdot k^2=k$ pieces of squares of side $k$)
$\displaystyle \quad  (1^{3}+2^{3}+3^{3}+\ldots+n^{3} )cm^3
\\= 1 \cdot (1^{2} cm^2\times 1 cm)+2 \cdot (2^{2}cm^2\times 1 cm)+3 \cdot (3^{2} cm^2\times 1cm)+\ldots+n \cdot (n^{2}cm^2\times 1 cm)
\\=(1cm+2cm+3cm+\cdots+ncm)^2\times 1cm$
Method 2(By Induction)
The statement is true for $n=1$, now assume it is true for some $n$, then $$
\begin{aligned}
\sum_{k=1}^{n+1} k^{3}&=\sum_{k=1}^{n} k^{3}+(n+1)^{3} \\
&=\left(\sum_{k=1}^{n} k\right)^{2}+n(n+1)^{2}+(n+1)^{2} \\
&=\left(\sum_{k=1}^{n} k\right)^{2}+2(n+1)\left(\sum_{k=1}^{n} k\right)+(n+1)^{2} \\
&=\left[\sum_{k=1}^{n} k+(n+1)\right]^{2} \\
&=\left(\sum_{k=1}^{n+1} k\right)^{2}
\end{aligned}
$$
Method 3(By difference)
Considering the difference yields
\begin{aligned}\\  n^{3}-(n-1)^{3}&=3 n^{2}-3 n+1 \\
\therefore  \sum_{k=1}^{n}\left[n^{3}-(n-1)^{3}\right]&=3 \sum_{k=1}^{n} n^{2}-3 \sum_{k=1}^{n} n+n \\ k^{4}-(k-1)^{4}&=4 k^{3}-6 k^{2}+4 k-1 \end{aligned}
Summing up yields
\begin{aligned}
 \sum_{k=1}^{n}\left[k^{4}-(k-1)^{4}\right]&=4 \sum_{k=1}^{n} k^{3}-6 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k-n \\
n^{4}&=4 \sum_{k=1}^{n} k^{3}-6 \cdot \frac{n}{6} \cdot(n+1)(2 n+1)+4 \cdot \frac{n}{2}(n+1)-n \\
4 \sum_{k=1}^{n} k^{3}&=n^{4}+n \left( n+1\right)(2 n+1)-2 n(n+1)+n  =[n(n+1)]^{2} \\
\sum_{k=1}^{n} k^{3}&=\left[\frac{n(n+1)}{2}\right]^{2}=(1+2+3+\cdots+n)^{2}
\end{aligned}
Wish you enjoy the explanation!
