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Summing powers (Posted on 2004-07-01) Difficulty: 3 of 5
I thought of three numbers.
Their sum is 6.
The sum of their squares is 8.
The sum of their cubes is 5.
What is the sum of their fourth powers?

See The Solution Submitted by Federico Kereki    
Rating: 3.0000 (5 votes)

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Solution An Alternative Methodology | Comment 16 of 18 |
(In reply to Answer by K Sengupta)

Let the numbers be x, y and z

Then, we have:

x+y+z = 6
x^2+ y^2 + z^2 = 8
x^3+y^3+z^3 = 5

Now;
x^2+y^2+z^2
=(x+y+z)^2 + 2t, where t = xy+yz+zx
or, t = (6^2 - 8)/2 = 14

Again,
x^3 + y^3 + z^3 - 3xyz
=(x+y+z)(x^2 + y^2+ z^2 - t)
or, 5 - 3xyz = 6(8-14)
Or, xyz = (5+36)/3 = 41/3

Again:

(xy+yz+zx)^2
= u + 2xyz(x+y+z), where u = x^2*y^2 + y^2*z^2 + z^2*x^2
Or, u = 14^2 - 2*(41/3)*6 = 32

Consequently,
x^4 + y^4 + z^4
= (x^2+y^2+z^2)^2 - 2u
= 64 - 2*32
= 0

Edited on May 31, 2007, 5:49 am
  Posted by K Sengupta on 2007-05-31 05:44:21

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