Let a,b,c,d be real numbers such that (b−c)(d−a)(a−b)(c−d)=52.Find the sum of all possible values of (a−b)(c−d)(a−c)(b−d).
Solution — click to reveal
From the given equation, 5(a−b)(c−d)=2(b−c)(d−a), which expands as 5ac−5ad−5bc+5bd=2bd−2ab−2cd+2ac.This simplifies to 2ab+3ac+3bd+2cd=5ad+5bc, so ad+bc=52ab+3ac+3bd+2cd.Then (a−b)(c−d)(a−c)(b−d)=ac−ad−bc+bdab−ad−bc+cd=ac+bd−52ab+3ac+3bd+2cdab+cd−52ab+3ac+3bd+2cd=5ac+5bd−2ab−3ac−3bd−2cd5ab+5cd−2ab−3ac−3bd−2cd=−2ab+2ac+2bd−2cd3ab−3ac−3bd+3cd=−2(ab−ac−bd+cd)3(ab−ac−bd+cd)=−23.