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 Quintic Expression (Posted on 2013-10-09)
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)
```

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 7 of 8 |
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)

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

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