Algebraic Proof on equivalence $$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$
Demonstrate the identities above considering a, b & c real numbers and distinct from each other.
I'm stucked on this problem. Take a look on what I did:
$x=(a-b)\therefore -x=(b-a)$;  $y=(a-c);-y=(c-a)$;  $z=(b-c);-z=(c-b)$.
___________________________________________________________________$$\frac{a^3}{xy}+\frac{b^3}{(-x)z}+\frac{c^3}{(-y)(-z)}$$
$$\frac{a^3z-b^3y+c^3x}{xyz}$$
$$\frac{a^3(b-c)+b^3(a-c)+c^3(a-b)}{xyz}$$
$$\frac{a^3(b-c)+b^3(\color{red}{-b}+a-c\color{red}{+b})+c^3(a-b)}{xyz}$$
$$\frac{a^3b+ab^3-a^3c+ac^3-b^3c-bc^3}{xyz}$$
$$\frac{ab(a^2+b^2)-ac(a^2-c^2)-bc(b^2+c^2)}{xyz}$$
after this step I was thinking about doing the same technique on the second term, $-ac(\color{red}{b^2}+a^2-c^2\color{red}{-b^2})$, but after doing some more  algebraic factorization I got stucked, could you help me to demonstrate?

Assuming $b\neq c$ the function
$$ f(z) = \frac{z^3}{(z-b)(z-c)}+\frac{b^3}{(b-c)(b-z)}+\frac{c^3}{(c-z)(c-b)} $$
is holomorphic over $\mathbb{C}\setminus\{b,c\}$. In the worst scenario $b$ and/or $c$ are simple poles, but $\operatorname*{Res}_{z=c}f(z)=\operatorname*{Res}_{z=b}f(z)=0$ imply that $f$ is entire (once its removable discontinuities are removed). Given that, by inspecting the definition above it is clear that $f(z)$ is a polynomial with degree $\leq 1$, and actually a linear polynomial of the form $f(z)=z+q$. Since
$$ f(0) = \frac{b^2}{b-c}+\frac{c^2}{c-b} = \frac{b^2-c^2}{b-c} = b+c $$
we have $f(z)=z+b+c$ for any $z\in\mathbb{C}\setminus\{b,c\}$, in particular for $z=a$.
