Sorry, but no. First, "just switch a and b" is clearly wrong, and we had multiple people propose that (Without the adjustment of sign). So not that trivial, I would argue. At least not more trivial than the underlying equality one wants to prove in the first place.

Let me fully spell out the proof that generality is not lost

Assume that b > a.

1. Swap the names a and b.

b^2-a^2 ?= (a+b)(b-a)

2. Multiply both sides by minus one

a^2-b^2 ?= -(a+b)(b-a)

3. Absorb the minus into the second factor

a^2-b^2 ?= (a+b)(a-b)

This is the same as the original equality we wanted to prove.

Now compare that to the purely algebraic proof for the whole theorem.

1. Distribute over the first parenthesis

(a+b)(b-a) = a(a-b) + b(a-b)

2. Distribute again

(a+b)(b-a) = a^2-ab + ba-b^2

3. Cancel equal terms

(a+b)(b-a) = a^2 - b^2

The proof that generality is not lost is of similar length to a full proof of the theorem!

So if you think the former can be skipped, then you must accept a proof of the whole theorem that simply reads "Follows from trivial algebraic manipulation"

They may be equally long, but length when written is IMO not a good measure of complexity.

The 2nd I would need pen and paper for to keep my head straight doing the distributive law. The 1st I can do in my head as:

"If b is larger than a, the left side will be negative instead of positive, but also "b-a" gets a negative sign, and these cancel each other, so it is the same"

But I agree with you anyway ...this is indeed "doing algebra".

I agree that the figure would have been better if it said that in that figure it displays the case where a is larger than b. And we should probably call it visualization of an algebraic proof, instead of visual proof.

The entire proof is trivial.