We have `f=X^3-3X^2+1`;How to find `x_1^(-4)+x_2^(-4)+x_3^(-4)`?; `x_1,x_2,x_3` are roots of `f(x)=0`.

We have:

`f(x) = x^3-3x^2+1`

`color(white)(f(x)) = (x-x_1)(x-x_2)(x-x_3)`

`color(white)(f(x)) = x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3`

So:

`{(x_1+x_2+x_3 = 3), (x_1x_2+x_2x_3+x_3x_1 = 0), (x_1x_2x_3 = -1):}`

Note that:

`x_1^(-4)+x_2^(-4)+x_3^(-4)`

can be written as a rational function of symmetric polynomials:

`x_1^(-4)+x_2^(-4)+x_3^(-4)=(x_2^4x_3^4+x_3^4x_1^4+x_1^4x_2^4)/(x_1^4x_2^4x_3^4)`

Then:

`x_2^4x_3^4+x_3^4x_1^4+x_1^4x_2^4" "` and `" "x_1^4x_2^4x_3^4`

are expressible in terms of the elementary symmetric polynomials (whose values we already know):

`{(x_1+x_2+x_3 = 3), (x_1x_2+x_2x_3+x_3x_1 = 0), (x_1x_2x_3 = -1):}`

We find:

`x_1^2+x_2^2+x_3^2 = (x_1+x_2+x_3)^2-2(x_1x_2+x_2x_3+x_3x_1)`

`color(white)(x_1^2+x_2^2+x_3^2) = color(blue)(3)^2-2(color(blue)(0))`

`color(white)(x_1^2+x_2^2+x_3^2) = color(green)(9)`

`x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2 = (x_1x_2+x_2x_3+x_3x_1)^2-2(x_1+x_2+x_3)x_1x_2x_3`

`color(white)(x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2) = (color(blue)(0))^2-2(color(blue)(3))(color(blue)(-1))`

`color(white)(x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2) = color(green)(6)`

`x_1^2x_2^2x_3^2 = (x_1x_2x_3)^2 = (color(blue)(-1))^2 = color(green)(1)`

Then:

`x_2^4x_3^4+x_3^4x_1^4+x_1^4x_2^4`

`=(x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2)^2 - 2(x_1^2+x_2^2+x_3^2)(x_1^2x_2^2x_3^2)`

`=(color(green)(6))^2-2(color(green)(9))(color(green)(1))`

`=36-18`

`=color(red)(18)`

`x_1^4x_2^4x_3^4 = (x_1^2x_2^2x_3^2)^2 = (color(green)(1))^2 = color(red)(1)`

So:

`x_1^(-4)+x_2^(-4)+x_3^(-4)=color(red)(18)/color(red)(1) = 18`

From `f` we have

`{(x_1+x_2+x_3=-3),(x_1x_2+x_1x_2+x_2x_3=0),(x_1x_2x_3=1):}`

Calling `y_1=1/x_1,y_2=1/x_2,y_3=1/x_3` we have

`{(y_1+y_2+y_3=(x_1x_2+x_1x_2+x_2x_3)/(x_1x_2x_3)=0),(y_1y_2+y_1y_3+y_2y_3=(x_1+x_2+x_3)/(x_1x_2x_3) =3 ),(y_1y_2y_3=1/(x_1x_2x_3)=1):}`

so the polynomial

`Y^3+3Y+1`

has as roots, the inverses of `f`'s

Now taking

`(y_1+y_2+y_3)^2 = y_1^2+y_2^2+y_3^2+2(y_1y_2+y_1y_3+y_2y_3)`

or

`0=y_1^2+y_2^2+y_3^2+2y_1y_2y_3(x_1+x_2+x_3)`

so

`y_1^2+y_2^2+y_3^2-6=0`

or

`y_1^2+y_2^2+y_3^2=6` and now

`(y_1^2+y_2^2+y_3^2)^2=y_1^4+y_2^4+y_3^4+2(y_1^2y_2^2+y_1^2y_3^2+y_2^2y_3^2)`

or

`36 = y_1^4+y_2^4+y_3^4+2y_1^2y_2^2y_3^2(x_1^2+x_2^2+x_3^2)`

but `x_1^2+x_2^2+x_3^2=9` and `y_1^2y_2^2y_3^2=1`

so finally

`36 = y_1^4+y_2^4+y_3^4+18` an then

`y_1^4+y_2^4+y_3^4 = 18 = 1/x_1^4+1/x_2^4+1/x_3^4`