Use Lagrange Interpolation polynomial to find this $\sum_{cyc}\frac{x^3}{(x^2-y^2)(x^2-z^2)}$ 
Let $x,y,z$ be the solutions of the equation $t^3-t^2+2t-3=0$. Find the sum
  $$\dfrac{x^3}{(x^2-y^2)(x^2-z^2)}+\dfrac{y^3}{(y^2-x^2)(y^2-z^2)}+\dfrac{z^3}{(z^2-x^2)(z^2-y^2)}$$

How can I use the Lagrange
interpolation polynomial to solve this question?
I did a lot of calculations to find this answer
$$\dfrac{x^3}{(x^2-y^2)(x^2-z^2)}+\dfrac{y^3}{(y^2-x^2)(y^2-z^2)}+\dfrac{z^3}{(z^2-x^2)(z^2-y^2)}=\dfrac{xy+yz+xz}{(x+y+z)(xy+yz+xz)-xyz}=\dfrac{2}{1\cdot 2+3}=\dfrac{2}{5}$$
I think we can use the Lagrange polynomial method to solve this problem?
 A: Here are some thoughts about the problem where I discuss how one would go about if one wants to use the Lagrange interpolation method and some obstacles to applying it.  I also give an example of a related problem where the method can easily be applied. I don't consider this a full answer, but it's too long for a comment and might be useful. 
Added: It turns out the method can indeed be applied. In the other answer Batominovski found a way to apply the method by first using the Vieta formulas to evaluate the troublesome terms $\frac{1}{(x+y)(x+z)}$ mentioned below and at the same time keeping the interpolation-polynomial quadratic.

You only have direct information about the three points $x,y,z$ plus the sum is also just three terms so if you want to contruct a Lagrange polynomial of some function to solve this problem it would lead to a quadratic polynomial. This interpolation formula would only be an identity (which we need it to be) if the function you try to interpolate is also at most quadratic. Thus in order to be able to easily apply Lagrange's formula we should be able to write the terms in the cyclic sum as $$\frac{x^3}{(x^2-y^2)(x^2-z^2)} = f(x)\frac{(t-y)(t-z)}{(x-y)(x-z)}$$
and similar for $y,z$ where $f$ is a quadratic polynomial and $t$ is the same in all three equations. In addition we need to be able to compute the numerical value of $f(t)$ so $t$ should preferably be a symmetric polynomial in $x,y,z$ so that we can use Vieta's formulas to do this. Since $x^3 = x^2-2x+3$ we can make $f(x)$ quadratic, but the remainding term $\frac{1}{(x+z)(x+y)}$ is not a symmetric function in the variables $x,y,z$ so I don't see a way to do this here. I cannot say for sure that it can't be done by performing some clever manipulations first, but at least there is no obvious application of the method you want for this problem so applying algebraic manipulations to put the sum in the form you have done and using Vieta's formulas is probably the best way to go.

I just want to mention that the method you want to use can indeed be applied for some related problems of this kind if the summand has the right properties and in these cases this significantly simplifies the solution. For example if your problem was slightly modified so that you instead were asked to find the sum
$$\sum_{\rm cyc}x^3\frac{(x+y)(x+z)}{(x-y)(x-z)}$$
where $x,y,z$ are roots of your equation $t^3-t^2+2t-3=0$ then this is possible. Using $t^2-2t+3 = t^3$ for $x,y,z$ the sum can be rewritten as
$$\left.\sum_{\rm cyc}(x^2-2x+3)\frac{(t-y)(t-z)}{(x-y)(x-z)}\right|_{t=x+y+z}$$
which is the Lagrange interpolation formula, evaluated at $t=x+y+z$, for the points $\{x,f(x)\},\{y,f(y)\},\{z,f(z)\}$ where $f(t) = t^2-2t+3$. Since $f$ is a quadratic polynomial the interpolation formula is an identity and it follows that
$$\sum_{\rm cyc}x^3\frac{(x+y)(x+z)}{(x-y)(x-z)} = f(x+y+z) = f(1) = 2$$
For this example we can also replace $x^3$ with any qubic polynomial $ax^3+bx^2+cx+d$ and still be able to use this method to evaluate the sum ($=2a+b+c+d$).
A: Yes, with some modification, we can use the Lagrange interpolation method.  First, note that $(y+z)(z+x)(x+y)=(x+y+z)(yz+zx+xy)-xyz$.  As $x+y+z=1$, $yz+zx+xy=2$, and $xyz=3$, $(y+z)(z+x)(x+y)=1\cdot 2-3=-1$.  The required expression is then
$$\begin{align}
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}
=\left(\frac{1}{(y+z)(z+x)(x+y)}\right)^2\,\sum_\text{cyc}x^3(y+z)^2\left(\frac{(x+z)(x+y)}{\left(x-y\right)\left(x-z\right)}\right)\,.
\end{align}$$
As $y+z=1-x$, we get $x^3(y+z)^2=x^3(1-x)^2=x^3-2x^4+x^5$.  Since $x^3-x^2+2x-3=0$, we obtain $x^3=x^2-2x+3$, $x^4=x^3-2x^2+3x=\left(x^2-2x+3\right)-2x^2+3x=-x^2+x+3$, and $x^5=x^4-2x^3+3x^2=\left(-x^2+x+3\right)-2\left(x^2-2x+3\right)+3x^2=5x-3$.  Hence, $$x^3(y+z)^2=
\left(x^2-2x+3\right)-2\left(-x^2+x+3\right)+(5x-3)=3x^2+x-6\,.$$
Similarly, $y^3(z+x)^2=3y^2+y-6$ and $z^3(x+y)^2=3z^2+z-6$.  That is, the quadratic polynomial $P(t):=3t^2+t-6$ interpolates the points $\left(x,x^3(y+z)^2\right)$, $\left(y,y^3(z+x)^2\right)$, and $\left(z,z^3(x+y)^2\right)$.  Ergo,
$$P(t)=\sum_\text{cyc}\,P(x)\left(\frac{(t-y)(t-z)}{(x-y)(x-z)}\right)\,.$$  That is,
$$\begin{align}
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}
&=\left(-1\right)^2\,\sum_\text{cyc}P(x)\left(\frac{(x+z)(x+y)}{\left(x-y\right)\left(x-z\right)}\right)
\\
&=\sum_\text{cyc}\,P(x)\left(\left.\frac{(t-y)(t-z)}{(x-y)(x-z)}\right|_{t=x+y+z}\right)
\\
&=\big.P(t)\big|_{t=x+y+z}=P(x+y+z)=P(1)=3\cdot 1^2+1-6=-2\,.
\end{align}$$
Your result is wrong, by the way.  (I have also verified my answer with Mathematica.)
In general, let $K$ be a field.  The cubic $t^3+at^2+bt+c$ in $K[t]$, where $a,b,c\in K$, has three pairwise distinct roots $x,y,z$ such that the sum of any two of them is nonzero, then we have $ab- c\neq 0_K$, $a^2b^2-4b^2-4a^3c+18abc-27c^2\neq 0_K$, and
$$
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}=-\frac{b}{ab-c}=\frac{yz+zx+xy}{(x+y+z)(yz+zx+xy)-xyz}\,.$$
In this case, $\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}=\frac{P(-a)}{(ab-c)^2}$, where $P(t):=-ct^2-(ac-b^2)t+bc$.
