Evaluating an expression given values of symmetric polynomials 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.   $33/60$
 A: 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 $(1)$
$$= \frac{50}{-60}$$
$$= \frac{-5}6$$
A: 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.
A: \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}
A: 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 $-\dfrac{4}{15}-\dfrac{3}{20}-\dfrac{5}{12} = -\dfrac{5}{6}$.
A: 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.
