Regarding the problem to evaluate:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

given:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

you might be wondering if the proposed solution is the only way to stab at the problem, or more specifically, if you solved it differently, but in a more tedious or long way, can you be proud of it?

Yes, indeed, in our book, regarding solving challenging and hard mathematics problems, solving it is what matters! You should be proud of yourself if you are capable and able to solve the hard mathematics problems! But, one should not be content with themselves, one should always hunger for knowledge and all other heuristic skills that would help greatly to become a better and creative problem solver!

Everyday is a chance to learn something, no, make it a plural, to learn many things so we are better than yesterday.

We promise you to provide light for the dark way and strength for the day and guidance for the hard and challenging problems!

Okay, back to the question, what if we approach it differently than the previous method? What if I make use of the information :

$xyz=4\,\,\rightarrow\,\,xy=\dfrac{4}{z},\,yz=\dfrac{4}{x},\,xz=\dfrac{4}{y}$

and rewrite the expression as:

[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]

Stop momentarily for now, to reflect. Always stop to reflect, never underestimate the power of your reflection, because when we do so, we tend to see things more clearly.

Okay, for the expression ** above, if we interpret it as the sum of the cubic equation with roots:

[MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1},\,\frac{1}{\dfrac{4}{z}+z-1}[/MATH]

and if we have the cubic equation with those roots as, for example, $m^3-6m^2+m-8=0$, then we can conclude that the desired expression as expressed in ** takes the value of 6.

Why is that? I will explain it below, from A to Z, just that you must bear with me to read the lengthy explanation...

First, note that the given information can lead us to write the cubic polynomial with the roots of $x,\,y$ and $z$.

I will provide a more general case here to avoid any potential convoluted confusion, note that if we have the roots of the cubic function $f(x)$ as $x,\,y$ and $z$, then we can work this cubic equation for $f(x)$ out by doing the following:

[MATH]f(m)=(m-x)(m-y)(m-z)[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=(m^2-xm-ym+xy)(m-z)[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-xm^2-ym^2-zm^2+ym+xm+zm-xyz[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-(x+y+z)m^2+(xy+yz+xz)m-xyz[/MATH]

Therefore, if we have the value for $x+y+z,\,\,xy+yz+xz$ and $xyz$ but not the values for each $x,\,y$ and $z$, we can still come up with the equation of the cubic polynomial with its roots as $x,\,y$ and $z$.

But, what if we are only given the information where:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH],

you might feel a bit upset, we don't have the value for $xy+yz+xz$, we have to work that out and when everything is still not clear to you if you are heading down the correct path, this might be a frustrating matter.

Don't fret, one really should be very familiar with the identity below:

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$

In other words, if we have $x+y+z$ and $x^2+y^2+z^2$, then the value of $xy+yz+xz$ can be obtained easily:

$xy+yz+xz=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}$

So, from [MATH]x+y+z=2,\,x^2+y^2+z^2=3[/MATH], we find indirectly that [MATH]xy+yz+zx=\dfrac{2^2-3}{2}=\dfrac{1}{2}[/MATH]

Now, we have all that is required to build the cubic equation with $x,\,y$ and $z$ as its roots, as we now obtain

[MATH]x+y+z=2,\,xy+yz+zx=\dfrac{1}{2}[/MATH] and [MATH]xyz=4[/MATH]

Therefore:

$m^3-(x+y+z)m^2+(xy+yz+zx)m-xyz=0$

[MATH]m^3-2m^2+\dfrac{m}{2}-4=0[/MATH]

[MATH]2m^3-4m^2+m-8=0[/MATH]

From:

[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]

Note that if we have another cubic equation with the roots defined as [MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1}[/MATH] and [MATH]\frac{1}{\dfrac{4}{z}+z-1}[/MATH], then [MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}[/MATH] is actually the sum of the roots, i.e. the value of the coefficient of the second term of the cubic equation writing in descending power of the exponents.

Now, 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 other words,

[MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}=-\dfrac{2}{9}[/MATH]

This actually begs the question, how on earth do we get the cubic equation $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$ where its roots are $k=\dfrac{1}{\dfrac{4}{x}+x-1},\,\dfrac{1}{\dfrac{4}{y}+y-1},\,\dfrac{1}{\dfrac{4}{z}+z-1}$?

Okay, we understand of your confusion and please stay tuned as we will guide you through it in the next post. :D

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

given:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

you might be wondering if the proposed solution is the only way to stab at the problem, or more specifically, if you solved it differently, but in a more tedious or long way, can you be proud of it?

Yes, indeed, in our book, regarding solving challenging and hard mathematics problems, solving it is what matters! You should be proud of yourself if you are capable and able to solve the hard mathematics problems! But, one should not be content with themselves, one should always hunger for knowledge and all other heuristic skills that would help greatly to become a better and creative problem solver!

Everyday is a chance to learn something, no, make it a plural, to learn many things so we are better than yesterday.

We promise you to provide light for the dark way and strength for the day and guidance for the hard and challenging problems!

Okay, back to the question, what if we approach it differently than the previous method? What if I make use of the information :

$xyz=4\,\,\rightarrow\,\,xy=\dfrac{4}{z},\,yz=\dfrac{4}{x},\,xz=\dfrac{4}{y}$

and rewrite the expression as:

[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]

Stop momentarily for now, to reflect. Always stop to reflect, never underestimate the power of your reflection, because when we do so, we tend to see things more clearly.

Okay, for the expression ** above, if we interpret it as the sum of the cubic equation with roots:

[MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1},\,\frac{1}{\dfrac{4}{z}+z-1}[/MATH]

and if we have the cubic equation with those roots as, for example, $m^3-6m^2+m-8=0$, then we can conclude that the desired expression as expressed in ** takes the value of 6.

Why is that? I will explain it below, from A to Z, just that you must bear with me to read the lengthy explanation...

First, note that the given information can lead us to write the cubic polynomial with the roots of $x,\,y$ and $z$.

I will provide a more general case here to avoid any potential convoluted confusion, note that if we have the roots of the cubic function $f(x)$ as $x,\,y$ and $z$, then we can work this cubic equation for $f(x)$ out by doing the following:

[MATH]f(m)=(m-x)(m-y)(m-z)[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=(m^2-xm-ym+xy)(m-z)[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-xm^2-ym^2-zm^2+ym+xm+zm-xyz[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-(x+y+z)m^2+(xy+yz+xz)m-xyz[/MATH]

Therefore, if we have the value for $x+y+z,\,\,xy+yz+xz$ and $xyz$ but not the values for each $x,\,y$ and $z$, we can still come up with the equation of the cubic polynomial with its roots as $x,\,y$ and $z$.

But, what if we are only given the information where:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH],

you might feel a bit upset, we don't have the value for $xy+yz+xz$, we have to work that out and when everything is still not clear to you if you are heading down the correct path, this might be a frustrating matter.

Don't fret, one really should be very familiar with the identity below:

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$

In other words, if we have $x+y+z$ and $x^2+y^2+z^2$, then the value of $xy+yz+xz$ can be obtained easily:

$xy+yz+xz=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}$

So, from [MATH]x+y+z=2,\,x^2+y^2+z^2=3[/MATH], we find indirectly that [MATH]xy+yz+zx=\dfrac{2^2-3}{2}=\dfrac{1}{2}[/MATH]

Now, we have all that is required to build the cubic equation with $x,\,y$ and $z$ as its roots, as we now obtain

[MATH]x+y+z=2,\,xy+yz+zx=\dfrac{1}{2}[/MATH] and [MATH]xyz=4[/MATH]

Therefore:

$m^3-(x+y+z)m^2+(xy+yz+zx)m-xyz=0$

[MATH]m^3-2m^2+\dfrac{m}{2}-4=0[/MATH]

[MATH]2m^3-4m^2+m-8=0[/MATH]

From:

[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]

Note that if we have another cubic equation with the roots defined as [MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1}[/MATH] and [MATH]\frac{1}{\dfrac{4}{z}+z-1}[/MATH], then [MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}[/MATH] is actually the sum of the roots, i.e. the value of the coefficient of the second term of the cubic equation writing in descending power of the exponents.

Now, 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 other words,

[MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}=-\dfrac{2}{9}[/MATH]

This actually begs the question, how on earth do we get the cubic equation $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$ where its roots are $k=\dfrac{1}{\dfrac{4}{x}+x-1},\,\dfrac{1}{\dfrac{4}{y}+y-1},\,\dfrac{1}{\dfrac{4}{z}+z-1}$?

Okay, we understand of your confusion and please stay tuned as we will guide you through it in the next post. :D

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