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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 Solution To Sum Of Sixth Powers | Comment 17 of 18 |
(In reply to An Alternative Methodology by K Sengupta)

x^6 + y^6 + z^6 - 3*(xyz)^2
= (x^2 + y^2 + z^2)[(x^4+y^4+z^4) - u], where :
u = x^2*y^2 + y^2*z^2 + z^2*x^2 = 32, as was shown in terms
of the previous post

Thus, x^6 + y^6 + z^6 - 3*(xyz)^2
= 8*(-32)
= -256, giving:

x^6+ y^6+ z^6 = 1681/3 - 256 = 913/3.

Edited on May 31, 2007, 5:51 am
  Posted by K Sengupta on 2007-05-31 05:47:12

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