Question
If x+y+z=1, x^2+y^2+z^2=2, x^3+y^3+z^3=3,
find the value of x^4+y^4+z^4
Answer
Given
x+y+z=1
(1)
x^2+y^2+z^2=2
(2)
x^3+y^3+z^3=3
(3)
xy+yz+zx = \dfrac{1}{2}[(x+y+z)^2-x^2-y^2-z^2]
(4)
Substituting (1) and (2) to (6) gives
xy+yz+zx =\dfrac{1}{2}(1-2) = -\dfrac{1}{2}
(5)
x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
(6)
Substituting (1),(2), (3) to (4) gives
3 -3xyz = 2-xy-yz-zx
3xyz = 1+xy+yz+zx
(7)
Substituting (5) to (7) gives
xyz = \dfrac{1}{6}
(8)
x^4+y^4+z^4
=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)
=(x^2+y^2+z^2)^2-2[(xy+yz+zx)^2-2xyz(x+y+z)]
(9)
Substituting (1),(2),(5), (8) to (9) gives
x^4+y^4+z^4
=2^2-2(\dfrac{1}{4}-2\cdot \dfrac{1}{6} )
=\dfrac{25}{6}