If# x+y+z=1#,#x^2+y^2+z^2=2# and #x^3+y^3+z^3=3# then what's the value of #x^4+y^4+z^4#?

Given:

#{ (x+y+z=1), (x^2+y^2+z^2=2), (x^3+y^3+z^3=3) :}#

The elementary symmetric polynomials in #x#, #y# and #z# are: #x+y+z#, #xy+yz+zx# and #xyz#. Once we find these, we can construct any symmetric polynomial in #x#, #y# and #z#. We are given #x+y+z#, so we just need to derive the other two...

Note that:

#2(xy+yz+zx) = (x+y+z)^2-(x^2+y^2+z^2) = -1#

So:

#xy+yz+zx = -1/2#

Note that:

#6xyz = (x+y+z)^3-3(x+y+z)(x^2+y^2+z^2)+2(x^3+y^3+z^3) = 1#

So:

#xyz = 1/6#

Then:

#x^4+y^4+z^4 = (x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)#

#color(white)(x^4+y^4+z^4) = (x^2+y^2+z^2)^2-2((xy+yz+zx)^2-2xyz(x+y+z))#

#color(white)(x^4+y^4+z^4) = 2^2-2((-1/2)^2-2(1/6)(1))#

#color(white)(x^4+y^4+z^4) = 4-2(1/4-1/3)#

#color(white)(x^4+y^4+z^4) = 4+2/12#

#color(white)(x^4+y^4+z^4)= 4+1/6#

#color(white)(x^4+y^4+z^4)= 25/6#

#color(white)()#
Bonus

Note that:

#(t-x)(t-y)(t-z) = t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz#

So substituting the values for the elementary symmetric polynomials that we found, we find that #x#, #y# and #z# are the three roots of:

#t^3-t^2-1/2t-1/6 = 0#

or if you prefer:

#6t^3-6t^2-3t-1=0#

In theory we could solve this using Cardano's method and directly evaluate #x^4+y^4+z^4#, but the methods used above are somewhat easier.

The three roots are:

#t_1 = 1/6(2 + root(3)(44 - 6 sqrt(26)) + root(3)(44 + 6 sqrt(26)))#

#t_2 = 1/6(2 + omega root(3)(44 - 6 sqrt(26)) + omega^2 root(3)(44 + 6 sqrt(26)))#

#t_3 = 1/6(2 + omega^2 root(3)(44 - 6 sqrt(26)) + omega root(3)(44 + 6 sqrt(26)))#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.