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Evaluate $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$

Given, $z+y+x=4, \qquad xyz=-60, \qquad xy+xz+yz=-17$

How do we do this? I found a common denominator, and substituted it for $-60$, but I am unaware of how to proceed.

Someone already asked the question but there is no useful answer. These are the possible answers:
A. $4/17$
B. $−5/6$
C. $17/60$
D. $−33/60$
E. $33/60$

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    $\begingroup$ I think you made a typo. I think it should be x/yz + y/xz +z/yx $\endgroup$ – 1-___- Jan 18 '17 at 18:39
  • $\begingroup$ For which values of $x$, $y$, and $z$ are you supposed to evaluate at? $\endgroup$ – Oiler Jan 18 '17 at 18:39
  • $\begingroup$ No its not a mistake, its a question from the UCL website, Here is the link, its question number 5 $\endgroup$ – exchangehelpforuni Jan 18 '17 at 18:41
  • $\begingroup$ ucl.ac.uk/clie/placement-tests/UPC/… $\endgroup$ – exchangehelpforuni Jan 18 '17 at 18:41
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    $\begingroup$ You have copied the problem from the site correctly, but I strongly suspect that the site itself made a mistake - the problem should be as XTL says. $\endgroup$ – Noah Schweber Jan 18 '17 at 18:46
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Assuming there is a typo.

We have to find -

\begin{align} & \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \\[10pt] = {} & \frac{x^2 + y^2 + z^2}{xyz} \tag 1 \end{align}

Also $x + y + z = 4$.

Squaring both sides,

$$x^2 + y^2 + z^2 + 2(xy + yz + zx) = 16$$

$$x^2 + y^2 + z^2 + 2(-17) = 16$$

$$x^2 + y^2 + z^2 = 16 + 34$$

$$x^2 + y^2 + z^2 = 50$$

Put this in equation $(1)$

$$= \frac{50}{-60}$$

$$= \frac{-5}6$$

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Hint: Since $xyz=-60$, we have $\frac{x}{yz}=-\frac{x^2}{60}$, and correspondingly for $\frac y{xz}$ and $\frac z{xy}$. Now calculate $(x+y+z)^2$ in two ways.

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Assuming that there is no typo I believe a possible solution is as follows:

Third equation can be written as:

$$ y(x+z)+xz=-17 $$

if we put $x+z=4-y$ obtained from the first equation and $xz=-60/y$ obtained from the second equation into the above equation we get

$$ y(4-y)-60/y=-17 $$

Above equation can be written as

$$ y^2(4-y)-60=-17y\Rightarrow y^3-4y^2-17y+60=0 $$

As can be seen from the above equation $x,y$ and $z$ are the roots of the above equation. (You can do the same operation for $x$ and $z$ and you get the above equation). By doing some trial above equation can be written as $$ y^3-4y^2-17y+60=0\Rightarrow(y-3)(y^2-y-20)=0\Rightarrow(y-3)(y-5)(y+4)=0 $$

So our $x,y$ and $z$ can be a tuple of 3,5 and -4. (The order doesn't matter as changing the order doesn't harm the given requirements).

However, from the asked quantity it follows that depending of the order of (3,5,-4) you can different results.(i.e. if (x,y,z)=(3,5,-4) you get -82/60 but if (x,y,z)=(-4,3,5) you get 75/60).

Thus I believe there is a problem with the question itself.

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Given the values of the symmetric polynomials, $x,y,z$ are the roots of:

$$t^3 - 4 t^2 -17 t + 60 = (t + 4)(t - 3)(t - 5) = 0$$

Therefore $\{x,y,z\} = \{-4,3,5\}\,$, but it's not possible to tell which value goes to which variable.

Since the expression $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$ is not symmetric in $x,y,z$ its value cannot be univocally determined. Therefore, as the other answers assumed, there is likely a typo in that expression.

Assuming $\dfrac x{yz} + \dfrac y {xz} + \dfrac z {xy}$, instead, the sum is $-\dfrac{4}{15}-\dfrac{3}{20}-\dfrac{5}{12} = -\dfrac{5}{6}$.

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\begin{align} & \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} = \frac{x^2 + y^2 + z^2}{xyz} \end{align}

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

so: \begin{align} \frac{x^2 + y^2 + z^2}{xyz} = \frac{(x+y+z)^2 - 2(xy + xz +yz)}{xyz} = \frac{16 - 2(-17)}{-60} = \frac{50}{-60} = -\frac{5}{6} \end{align}

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