Number of different value of given expression for $x,y,z \in \{1,2,3,4,5\}$ Let $x,y,z \in \{1,2,3,4,5\}$, then find number of different values of
$$\dfrac{x^3}{(x-z)(x-y)}+\dfrac{y^3}{(y-x)(y-z)}+\dfrac{z^3}{(z-x)(z-y)}$$
Could someone please share the approach to start this question. I tried to factorize the expression after taking L.C.M. but the whole gets very messy. How should I start?
 A: First, by observing the form of the denominator, we can immediately pose the condition that $x \neq y \neq z$ for the fraction to be well-defined.
Next, because all $3$ fractions are identical, permutation of a set of $(x,y,z)$ will all result in the same value (i.e. $(x,y,z) = (1,2,3)$ will give the same value as $(x,y,z) = (2,3,1)$ and so on).
With this, we narrow down the $125$ possibilities of $(x,y,z)$ to ${5 \choose 3} = 10$ possibilities; namely
$$
(1,2,3)\quad (1,2,4)\quad (1,2,5)\quad (1,3,4)\quad (1,3,5) \\
(1,4,5)\quad (2,3,4)\quad (2,3,5)\quad (2,4,5)\quad (3,4,5)
$$
(Optional) We can simplify the equation as follows:
\begin{align}
\frac{x^3}{(x-y)(x-z)} + \frac{y^3}{(y-x)(y-z)} + \frac{z^3}{(z-x) (z-y)}
&= \frac{x^3(y-z) - y^3(x-z) + z^3(x-y)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3(y-z) + yz(y^2 - z^2) - x(y^3 - z^3)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3(y-z) + yz(y-z)(y+z) - x(y-z)(y^2+yz+z^2)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3 + yz(y+z) - x(y^2+yz+z^2)}{(x-y)(x-z)} \\
&= \frac{x(x^2-y^2)-yz(x-y)-z^2(x-y)}{(x-y)(x-z)} \\
&= \frac{x(x+y)-yz-z^2}{x-z} \\
&= x+y+z
\end{align}
From here, just manually substitute these $10$ combinations and you will find that the possible values are $\{6,7,8,9,10,11,12\}$.
