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?

1 Answer
George C.
Jul 9, 2017

x^4+y^4+z^4 = 25/6

Explanation:

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.

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