### An Engaging Algebraic Identity

A question has been asked on a linkedin group to prove the following engaging identity

Prove

provided .

One of the posts pointed to a solution at Stevens Society of Mathematicians. What follows is a slight simplification of that proof.

Denote the left factor and the right factor . Observe that whenever two of the arguments in are equal, the whole expression vanishes. For example,

Adding the fractions in , . What we just showed implies that the numerator is divisible by . Multiplying through confirms that .

Now, let's turn to the right factor. Up to now we have not used the condition . It's time we do. Introduce

Seen as a system of linear equations with as unknown, it's degenerate because . The situation improves if we replace any of the equations with . Then, for example,

Similarly, and . This allows us to express the right factor in terms of :

This is exactly the same form as , implying that

.

Finally,