Cubes in Pascal's Triangle

Tony Foster discovered cubes in Pascal's triangle and rigorously established their presence.

$\displaystyle n^{3}=\bigg[C^{n+1}_{2}\cdot C^{n-1}_{1}\cdot C^{n}_{0}\bigg] + \bigg[C^{n+1}_{1}\cdot C^{n}_{2}\cdot C^{n-1}_{0}\bigg] + C^{n}_{1}.$

Star of David cubes in Pascal triangle

The verification is simple:

$\begin{align} \displaystyle C^{n+1}_{2}\cdot C^{n-1}_{1}\cdot C^{n}_{0}&=\frac{(n+1)n}{2}\cdot (n-1)\\ &=\frac{(n+1)n(n-1)}{2}. \end{align}$

And similarly,

$\begin{align} \displaystyle C^{n+1}_{1}\cdot C^{n}_{1}\cdot C^{n-1}_{0}&=(n+1)\cdot \frac{n(n-1)}{2}\cdot\\ &=\frac{(n+1)n(n-1)}{2}. \end{align}$

Summing up:

$\begin{align} 2\frac{(n+1)n(n-1)}{2}+C^{n}_{1}&=(n+1)n(n-1)+n\\ &=n(n^2-1)+n=n^3-n+n=n^3. \end{align}$

Tony also pointed to a related sighting of squares:

$n^{2}=\bigg[C^{n+1}_{2}-C^{n-1}_{0}\bigg]+\bigg[C^{n+1}_{1}-C^{n-1}_{1}\bigg]+\bigg[C^{n}_{2}-C^{n}_{0}\bigg].$

Verification is left as an exercise.

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