Let a,b,c and d be distinct real numbers such that

a+b+c+d=3
a²+b²+c²+d²=45

Find the value of the expression

       a5                b5                c5                d5
--------------- + --------------- + --------------- + --------------
(a-b)(a-c)(a-d)   (b-a)(b-c)(b-d)   (c-a)(c-b)(c-d)   (d-a)(d-b)(d-c)

Starting with the identity

a⁵(b-c)(b-d)(c-d) - b⁵(a-c)(a-d)(c-d) + c⁵(a-b)(a-d)(b-d) - d⁵(a-b)(a-c)(b-c)

= (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)[a²+b²+c²+d²+ab+ac+ad+bc+bd+cd]

and then dividing by (a-b)(a-c)(a-d)(b-c)(b-d)(c-d), for distinct a,b,c,d:

a⁵/(a-b)(a-c)(a-d)+b⁵/(b-a)(b-c)(b-d)+c⁵/(c-a)(c-b)(c-d)+d⁵/(d-a)(d-b)(d-c)

            = a² + b² + c² + d² + ab + ac + ad + bc + bd + cd

            =  a² + b² + c² + d² + [(a + b + c + d)² – (a² + b² + c² + d²)]/2

            = [a² + b² + c² + d² + (a + b + c + d)²]/2

            = [45 + 3²]/2     using the given equalities

            = 27

.. with a little help from Mathematica in confirming the initial identity.

Why the restriction to real numbers?

Posted by Harry on 2013-10-12 13:26:36