We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac.
Comparing (p – 5q – 3r)2 with (a + b + c)2 , we get:
a = p, b = –5q and c = –3r
Substituting these in the above formula:
(p – 5q – 3r)2 = (p)2 + (–5q)2 + (–3r)2 + 2 × p × (-5q) + 2 × (–5q) × (–3r) + 2 × p × (–3r)
= p2 + 25q2 + 9r2 – 10pq + 30qr – 6pr