Question

If `x+y+z=6` and `x^2+y^2+z^2=14` and `x^3+y^3+z^3=36` then what are `x, y, z` ?

Answer 1

`{x, y, z} = {1, 2, 3}`

Explanation:

`1+2+3 = 6`

`1^2+2^2+3^2 = 1+4+9 = 14`

`1^3+2^3+3^3 = 1+8+27 = 36`

So `{x, y, z} = {1, 2, 3}` is a solution - actually `3! = 6` solutions since any order of `x, y, z` is possible.

`color(white)()`
Bonus

If there was not a simple solution that could be guessed, how would you solve a system of equations of this form?

Suppose:

`{ (x+y+z = a), (x^2+y^2+z^2 = b), (x^3+y^3+z^3 = c) :}`

Note that:

`a^2-b = (x+y+z)^2-(x^2+y^2+z^2)`

`color(white)(a^2-b) = 2(xy+yz+zx)`

`ab-c=(x+y+z)(x^2+y^2+z^2) - (x^3+y^3+z^3)`

`color(white)(ab-c)= xy^2+yz^2+zx^2+x^2y+y^2z+z^2x`

`a^3-c = (x+y+z)^3 - (x^3+y^3+z^3)`

`color(white)(a^3-c) = 3(xy^2+yz^2+zx^2+x^2y+y^2z+z^2x)+6xyz`

`color(white)(a^3-c) = 3(ab-c)+6xyz`

So we have:

`{ (x+y+z = a), (xy+yz+zx = (a^2-b)/2), (xyz = (a^3-3ab+2c)/6) :}`

Then `x, y, z` are the roots of:

`0 = (t-x)(t-y)(t-z)`

`color(white)(0) = t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz`

`color(white)(0) = t^3-at^2+(a^2-b)/2t-(a^3-3ab+2c)/6`

Let's check that with the given example:

`a = 6, b=14, c=36`

`0 = t^3-color(blue)(6)t^2+(color(blue)(6)^2-color(blue)(14))/2t-(color(blue)(6)^3-3(color(blue)(6))(color(blue)(14))+2(color(blue)(36)))/6`

`color(white)(0) = t^3-6t^2+11t-6`

`color(white)(0) = (t-1)(t-2)(t-3)`