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#.

2 Answers
George C.
May 10, 2017

#x_1^(-4)+x_2^(-4)+x_3^(-4) = 18#

Explanation:

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#

Cesareo R.
May 11, 2017

See below.

Explanation:

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#

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