## Tuesday, April 7, 2015

### How to obtain the cubic polynomial from our previous problem?

In the previous problem's post ((2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1 ), we said if we let $k=\dfrac{1}{\dfrac{4}{x}+x-1}$, we will then have $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.

In this post, we will reveal the logic and reason behind it and guide you step-by-step in achieving the other cubic polynomial where $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.

Given that $x$ satisfies:

$2x^3-4x^2+x-8 = 0$

Now, if we let $k=\dfrac{1}{\dfrac{4}{x} + x-1}$ and allow for $k$ to satisfy the cubic:

$k^3 + ak^2 + bk + c = 0$, i.e.

$\left(\dfrac{1}{\dfrac{4}{x} + x-1}\right)^3 + a\left(\dfrac{1}{\dfrac{4}{x} + x-1}\right)^2 + b\left(\dfrac{1}{\dfrac{4}{x} + x-1}\right) + c = 0$

We see that we could simplify it and get:

[MATH]\color{yellow}\bbox[5px,green]{cx^6+(b-3c)x^5+(a-2b+15c)x^4+(1-a+9b-25c)x^3+(4a-8b+60c)x^2+(16b-48c)x+64c=0}[/MATH]

What are we going to do now, is to reduce it to a quadratic polynomial and therefore, we need to relate $x^6,\,x^5,\,x^4,\,x^3$ with $x^2$.

First recognizing that
$x^3=2x^2-\dfrac{x}{2}+4$

$x^4=x\left(2x^2-\dfrac{x}{2}+4\right)$

$\,\,\,\,\,\,\,\,=2x^3-\dfrac{x^2}{2}+4x$

$\,\,\,\,\,\,\,\,=2\left(2x^2-\dfrac{x}{2}+4\right)-\dfrac{x^2}{2}+4x$

$\,\,\,\,\,\,\,\,=\dfrac{7x^2}{2}+3x+8$

By the same token, we get

$x^5=10x^2+\dfrac{25x}{4}+3x+14$

$x^6=\dfrac{105x^2}{4}+9x+40$

Imposing the cubic in $x$ reduces this to:

[MATH]\color{yellow}\bbox[5px,green]{cx^6+(b-3c)x^5+(a-2b+15c)x^4+(1-a+9b-25c)x^3+(4a-8b+60c)x^2+(16b-48c)x+64c=0}[/MATH]

$\left( 13\,b+2+\frac{11}{2}\,a+{\frac {235}{4}}\,c \right) {x}^{2}+ \left( {\frac {47}{4}}\,b-\frac{1}{2}-\frac{1}{4}\,c+\frac{7}{2}\,a \right) x+4+34\,b+82\,c+4\,a = 0$

and setting the coefficients to zero:

$13\,b+2+\frac{11}{2}\,a+{\frac {235}{4}}\,c =0$

$\dfrac {47b}{4}-\dfrac{1}{2}-\dfrac{c}{4}+\dfrac{7a}{2}=0$

$4+34b+82c+4a=0$

and solving for $a$ gave us what we were looking for.

In fact, $(a,\,b,\,c)=(\dfrac{2}{9},\,-\dfrac{2}{81},\,-\dfrac{4}{9})$, and so this shows :

$k^3 +\dfrac{2}{9}k^2 -\dfrac{2}{81}k -\dfrac{4}{9} = 0$

Now rewriting it so that the coefficient of the term $x^2$ is preceded by a negative sign, we get:

$k^3 -\left(-\left(\dfrac{2}{9}\right)\right)k^2 -\dfrac{2}{81}k -\dfrac{4}{9} = 0$

This clearly shows that the sum of the three roots:

$\dfrac{1}{\dfrac{4}{x} + x-1},\,\dfrac{1}{\dfrac{4}{y} + y-1},\,\dfrac{1}{\dfrac{4}{z} + z-1}$

of the cubic equation:

$k^3 -\left(-\left(\dfrac{2}{9}\right)\right)k^2 -\dfrac{2}{81}k -\dfrac{4}{9} = 0$

is:

$-\dfrac{2}{9}$.