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- Feb 14, 2012

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- Feb 7, 2012

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The equation with roots $x,y,z$ is thus $t^3 - 2t^2 + \frac12t - 4 = 0$. (This equation only has one real root, so it's just as well that the question is about complex numbers.)

Now let $\alpha = yz+x$, $\beta = zx+y$ and $\gamma = xy+z$. The next step is to find the symmetric functions $\sum\alpha$, $\sum\beta\gamma$ and $\alpha\beta\gamma$. For this, we first need to find the value of $\sum x\bigl(y^2+z^2\bigr)$ and $\sum y^2z^2$. To do that, notice that $\sum x \sum yz = 3xyz + \sum x\bigl(y^2+z^2\bigr)$, from which $\sum x\bigl(y^2+z^2\bigr) = -11.$ Also, $\Bigl(\sum yz\Bigr)^2 = \sum y^2z^2 + 2xyz\sum x$, from which $\sum y^2z^2 = \frac14 - 16 = -\frac{63}4.$

We can now calculate that $$\textstyle \sum\alpha = \sum yz + \sum x = \tfrac72,$$ $$\textstyle \sum\beta\gamma = \sum\bigl(x^2yz + xy^2 + xz^2 + yz\bigr) = xyz\sum x + \sum x\bigl(y^2+z^2\bigr) + \sum yz = 8 - 11 + \tfrac12 = -\tfrac52,$$ $$\textstyle \alpha\beta\gamma = (yz+x)(zx+y)(xy+z) = (xyz)^2 + xyz\sum x^2 + \sum y^2z^2 + xyz = 16 + 12 - \tfrac{63}4 + 4 = \tfrac{65}4.$$

From those calculations, the equation with roots $\alpha,\,\beta,\,\gamma$ is $t^3 - \frac72t^2 - \frac52t - \frac{65}4$. The equation with roots $\alpha-1,\,\beta-1,\,\gamma-1$ is $(t+1)^3 - \frac72(t+1)^2 - \frac52(t+1) - \frac{65}4$, or $4t^3 -2t^2 -26t - 85 = 0.$ Replacing $t$ by $1/t$, it follows that the equation with roots $1/(\alpha-1),\,1/(\beta-1),\,1/(\gamma-1)$ is $85t^3 + 26t^2 + 2t - 4 = 0.$

Finally, the sum of the roots of that last equation is $$-\frac{26}{85} = \frac1{\alpha-1} + \frac1{\beta-1} + \frac1{\gamma-1} = \frac1{yz+x-1} +\frac1{zx+y-1} + \frac1{xy+z-1}.$$

Disclaimer: I have not checked the details, so the answer may be wrong. But the method should be correct.

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- Jan 26, 2012

- 183

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

Next, we can re-write our target equation using $xyz=4$ as

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

If we let $w = \dfrac{1}{\dfrac{4}{x} + x - 1}$ then we can show that

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

so the sum of the three roots, which is our target, is $-\dfrac{2}{9}$.

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- Feb 14, 2012

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I solved it differently and here is my solution:

\(\displaystyle =\frac{1}{xy+(2-x-y)-1}+\frac{1}{yz+(2-y-z)-1}+\frac{1}{xz+(2-x-z)-1}\)

\(\displaystyle =\frac{1}{xy-x-y+1}+\frac{1}{yz-y-z+1}+\frac{1}{xz-x-z+1}\)

\(\displaystyle =\frac{1}{(x-1)(y-1)}+\frac{1}{(y-1)(z-1)}+\frac{1}{(x-1)(z-1)}\)

\(\displaystyle =\frac{(z-1)+(x-1)+(y-1)}{(x-1)(y-1)(z-1)}\)

\(\displaystyle =\frac{2-3}{xyz-(xy+yz+xz)+(x+y+z)-1}\)

\(\displaystyle =\frac{-1}{4-\frac{1}{2}+(2)-1}\)

\(\displaystyle =-\frac{2}{9}\)

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- Feb 14, 2012

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I have tried many ways to figure out how could we end up with the cubic function of $ w^3 + \dfrac{2}{9}w^2 - \dfrac{2}{81}w - \dfrac{4}{81} = 0$ if we let $w = \dfrac{1}{\dfrac{4}{x} + x - 1}$...with no luck.

My convoluted attempt to generalize that cubic function is a great deal more difficult than to find the value for the target expression, LOL!

Could you please tell me how did you obtain $ w^3 + \dfrac{2}{9}w^2 - \dfrac{2}{81}w - \dfrac{4}{81} = 0$ as you showed in your solution?

Many, many thanks in advance.

- Jan 26, 2012

- 183

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

then it was a matter of tour de force

Letting $w =\dfrac{1}{\dfrac{4}{x} + x-1}$ and allowing that $w$ satisfy a cubic

$w^3 + aw^2 + bw + c = 0$

Simplifying gave

$c{x}^{6}+ \left( b-3\,c \right) {x}^{5}+ \left( -2\,b+a+15\,c \right)

{x}^{4}+ \left( 1-a+9\,b-25\,c \right) {x}^{3}$

$+ \left( -8\,b+4\,a+60\,

c \right) {x}^{2}+ \left( 16\,b-48\,c \right) x+64\,c=0$.

Imposing the cubic in $x$ reduced this to

$\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 and solving for $a, b$ and $c$ gave me what I was looking for.

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- Feb 14, 2012

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Hi **Jester**,

Great explanation and thanks for the insights!

Great explanation and thanks for the insights!