(p-q)2 + (q-p)2

Notice that (p - q)² = p² - 2pq + q²

But notice that (q - p)² is exactly the same result....(prove this for yourself)

So all we really need to do is to just double the first result, and we get

2p² - 4pq + 2q²

.....as Aziz  found !!!

(p-q)2 + (q-p)2

(p-q)(p-q) + (q-p)(q-p)

(p² - pq - pq + q²) + (q² - pq - pq + p²)

2p² - 4pq + 2q².

$$(p-q)^2$$ = $$p^2-2pq+q^2$$   This is the first one'...

Then;

$$(q-p)^2 =q^2-2pq+ p^2$$ This is second one..

After that,,

We must add the first  & second

$$(p-q)^2$$ + $$(q-p)^2$$ This is equal to that;

= $$p^2-2pq+q^2$$ $$+ q^2-2pq +p^2$$

= $$p^2+p^2+q^2+q^2-4pq$$

= $$2p^2+2q^2-2pq$$ That is the final answer.