find x^4 + y^4 + z^4 when

x+y+z=3
x^2+y^2+z^2=5
x^3+y^3+z^3=7

solution

because there are 3 variables let them be solution of

f(t) = t^3 + at^2 + bt + c = 0

sum of roots = 3 so a = - 3

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

t^3 - 3t^2 + 2t +c = 0

f(x) = x^3 - 3x^2 + 2x + c = 0
f(y) = y^3 - 3y^2 + 2y + c = 0
f(z) = z^3 - 3z^2 + 2 z + c = 0

so add to get (x^3 + y^3 + z^3) - 3(x^2 + y^2+ z^2) + 2 (x+y+z) + 3c = 0

or 7 - 3 * 5 + 2 * 3 + 3c = 0

or 3c = 2

so f(t) = t^3 - 3t^2 + 2t + 2/3 = 0

multiply by t to get

t^4 - 3 t^3 + 2t^2 + 2/3 t = 0

x,y ,z satisfy this and adding putting x, y, z for t and adding we get

x^4 + y^4 + z^4 = 3(x^3+y^3+z^3) - 2 (x^2 + y^2 + z^2 ) - 2/3(x+y+z)
= 3 * 7 - 2 * 5 - 2/3 * 3 = 21-10 - 2 = 9