<s>[INST] 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?
 [/INST] 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 $P(t):=-ct^2-(ac-b^2)t+bc$.
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