Please help me about a very hard equation!

May 2016
10
0
Istanbul
x+y+z=1
x^2+y^2+z^2=2
x^3+y^3+z^3=3
x^4+y^4+z^4=?

answer is not 4 :)

thanks.
 

romsek

MHF Helper
Nov 2013
6,838
3,079
California
$x^4+y^4+z^4=\dfrac{25}{6}$
 
Last edited by a moderator:
  • Like
Reactions: 1 person
May 2016
10
0
Istanbul
Thanks for your answer but I don't understand this. Could you describe in more detail? Thanks.
 
Feb 2015
2,255
510
Ottawa Ontario
x+y+z=1
x^2+y^2+z^2=2
1st equation doubled: 2(x + y + z) = 2
SO:
x^2 + y^2 + z^2 = 2x + 2y + 2z
Rearranged:
x^2 - 2x + y^2 + z^2 - 2y - 2z = 0
Let k = y^2 + z^2 - 2y - 2z; then:

x^2 - 2x + k = 0
Solve for x:
x = 1 +- SQRT(1 - k)

Poor x has very little chance of being real!!
 
  • Like
Reactions: 1 person
Jun 2013
1,152
615
Lebanon
express in terms of elementary symmetric functions. Let

\(\displaystyle x+y+z=s\)

\(\displaystyle x y + y z + z x = r\)

\(\displaystyle x y z=p\)

we get

\(\displaystyle x^2+y^2+z^2=s^2-2r=2\)

\(\displaystyle x^3+y^3+z^3=3p+s^3-3s r=3\)

since \(\displaystyle s=1\) the above equations give \(\displaystyle r=-1/2, p= 1/6\)

therefore

\(\displaystyle x^4+y^4+z^4=4p s+s^4-4s^2r+2r^2=\frac{25}{6}\)
 
  • Like
Reactions: 1 person