# 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.

May 10, 2017

${x}_{1}^{- 4} + {x}_{2}^{- 4} + {x}_{3}^{- 4} = 18$

#### Explanation:

We have:

$f \left(x\right) = {x}^{3} - 3 {x}^{2} + 1$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - \left({x}_{1} + {x}_{2} + {x}_{3}\right) {x}^{2} + \left({x}_{1} {x}_{2} + {x}_{2} {x}_{3} + {x}_{3} {x}_{1}\right) x - {x}_{1} {x}_{2} {x}_{3}$

So:

$\left\{\begin{matrix}{x}_{1} + {x}_{2} + {x}_{3} = 3 \\ {x}_{1} {x}_{2} + {x}_{2} {x}_{3} + {x}_{3} {x}_{1} = 0 \\ {x}_{1} {x}_{2} {x}_{3} = - 1\end{matrix}\right.$

Note that:

${x}_{1}^{- 4} + {x}_{2}^{- 4} + {x}_{3}^{- 4}$

can be written as a rational function of symmetric polynomials:

${x}_{1}^{- 4} + {x}_{2}^{- 4} + {x}_{3}^{- 4} = \frac{{x}_{2}^{4} {x}_{3}^{4} + {x}_{3}^{4} {x}_{1}^{4} + {x}_{1}^{4} {x}_{2}^{4}}{{x}_{1}^{4} {x}_{2}^{4} {x}_{3}^{4}}$

Then:

${x}_{2}^{4} {x}_{3}^{4} + {x}_{3}^{4} {x}_{1}^{4} + {x}_{1}^{4} {x}_{2}^{4} \text{ }$ and $\text{ } {x}_{1}^{4} {x}_{2}^{4} {x}_{3}^{4}$

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

$\left\{\begin{matrix}{x}_{1} + {x}_{2} + {x}_{3} = 3 \\ {x}_{1} {x}_{2} + {x}_{2} {x}_{3} + {x}_{3} {x}_{1} = 0 \\ {x}_{1} {x}_{2} {x}_{3} = - 1\end{matrix}\right.$

We find:

${x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} = {\left({x}_{1} + {x}_{2} + {x}_{3}\right)}^{2} - 2 \left({x}_{1} {x}_{2} + {x}_{2} {x}_{3} + {x}_{3} {x}_{1}\right)$

$\textcolor{w h i t e}{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}} = {\textcolor{b l u e}{3}}^{2} - 2 \left(\textcolor{b l u e}{0}\right)$

$\textcolor{w h i t e}{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}} = \textcolor{g r e e n}{9}$

${x}_{1}^{2} {x}_{2}^{2} + {x}_{2}^{2} {x}_{3}^{2} + {x}_{3}^{2} {x}_{1}^{2} = {\left({x}_{1} {x}_{2} + {x}_{2} {x}_{3} + {x}_{3} {x}_{1}\right)}^{2} - 2 \left({x}_{1} + {x}_{2} + {x}_{3}\right) {x}_{1} {x}_{2} {x}_{3}$

$\textcolor{w h i t e}{{x}_{1}^{2} {x}_{2}^{2} + {x}_{2}^{2} {x}_{3}^{2} + {x}_{3}^{2} {x}_{1}^{2}} = {\left(\textcolor{b l u e}{0}\right)}^{2} - 2 \left(\textcolor{b l u e}{3}\right) \left(\textcolor{b l u e}{- 1}\right)$

$\textcolor{w h i t e}{{x}_{1}^{2} {x}_{2}^{2} + {x}_{2}^{2} {x}_{3}^{2} + {x}_{3}^{2} {x}_{1}^{2}} = \textcolor{g r e e n}{6}$

${x}_{1}^{2} {x}_{2}^{2} {x}_{3}^{2} = {\left({x}_{1} {x}_{2} {x}_{3}\right)}^{2} = {\left(\textcolor{b l u e}{- 1}\right)}^{2} = \textcolor{g r e e n}{1}$

Then:

${x}_{2}^{4} {x}_{3}^{4} + {x}_{3}^{4} {x}_{1}^{4} + {x}_{1}^{4} {x}_{2}^{4}$

$= {\left({x}_{1}^{2} {x}_{2}^{2} + {x}_{2}^{2} {x}_{3}^{2} + {x}_{3}^{2} {x}_{1}^{2}\right)}^{2} - 2 \left({x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}\right) \left({x}_{1}^{2} {x}_{2}^{2} {x}_{3}^{2}\right)$

$= {\left(\textcolor{g r e e n}{6}\right)}^{2} - 2 \left(\textcolor{g r e e n}{9}\right) \left(\textcolor{g r e e n}{1}\right)$

$= 36 - 18$

$= \textcolor{red}{18}$

${x}_{1}^{4} {x}_{2}^{4} {x}_{3}^{4} = {\left({x}_{1}^{2} {x}_{2}^{2} {x}_{3}^{2}\right)}^{2} = {\left(\textcolor{g r e e n}{1}\right)}^{2} = \textcolor{red}{1}$

So:

${x}_{1}^{- 4} + {x}_{2}^{- 4} + {x}_{3}^{- 4} = \frac{\textcolor{red}{18}}{\textcolor{red}{1}} = 18$

May 11, 2017

See below.

#### Explanation:

From $f$ we have

$\left\{\begin{matrix}{x}_{1} + {x}_{2} + {x}_{3} = - 3 \\ {x}_{1} {x}_{2} + {x}_{1} {x}_{2} + {x}_{2} {x}_{3} = 0 \\ {x}_{1} {x}_{2} {x}_{3} = 1\end{matrix}\right.$

Calling ${y}_{1} = \frac{1}{x} _ 1 , {y}_{2} = \frac{1}{x} _ 2 , {y}_{3} = \frac{1}{x} _ 3$ we have

$\left\{\begin{matrix}{y}_{1} + {y}_{2} + {y}_{3} = \frac{{x}_{1} {x}_{2} + {x}_{1} {x}_{2} + {x}_{2} {x}_{3}}{{x}_{1} {x}_{2} {x}_{3}} = 0 \\ {y}_{1} {y}_{2} + {y}_{1} {y}_{3} + {y}_{2} {y}_{3} = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{{x}_{1} {x}_{2} {x}_{3}} = 3 \\ {y}_{1} {y}_{2} {y}_{3} = \frac{1}{{x}_{1} {x}_{2} {x}_{3}} = 1\end{matrix}\right.$

so the polynomial

${Y}^{3} + 3 Y + 1$

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

Now taking

${\left({y}_{1} + {y}_{2} + {y}_{3}\right)}^{2} = {y}_{1}^{2} + {y}_{2}^{2} + {y}_{3}^{2} + 2 \left({y}_{1} {y}_{2} + {y}_{1} {y}_{3} + {y}_{2} {y}_{3}\right)$

or

$0 = {y}_{1}^{2} + {y}_{2}^{2} + {y}_{3}^{2} + 2 {y}_{1} {y}_{2} {y}_{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right)$

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

${\left({y}_{1}^{2} + {y}_{2}^{2} + {y}_{3}^{2}\right)}^{2} = {y}_{1}^{4} + {y}_{2}^{4} + {y}_{3}^{4} + 2 \left({y}_{1}^{2} {y}_{2}^{2} + {y}_{1}^{2} {y}_{3}^{2} + {y}_{2}^{2} {y}_{3}^{2}\right)$

or

$36 = {y}_{1}^{4} + {y}_{2}^{4} + {y}_{3}^{4} + 2 {y}_{1}^{2} {y}_{2}^{2} {y}_{3}^{2} \left({x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2}\right)$

but ${x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} = 9$ and ${y}_{1}^{2} {y}_{2}^{2} {y}_{3}^{2} = 1$

so finally

$36 = {y}_{1}^{4} + {y}_{2}^{4} + {y}_{3}^{4} + 18$ an then

${y}_{1}^{4} + {y}_{2}^{4} + {y}_{3}^{4} = 18 = \frac{1}{x} _ {1}^{4} + \frac{1}{x} _ {2}^{4} + \frac{1}{x} _ {3}^{4}$