Cyclic Equation. Prove that: $\small\frac { a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3 }{ (a-b)(b-c)(c-a) } = ab + bc + ca$? This is how far I got without using polynomial division:
\begin{align}
\tiny
\frac { a^{ 2 }(b-c)^{ 3 }+b^{ 2 }(c-a)^{ 3 }+c^{ 2 }(a-b)^{ 3 } }{ (a-b)(b-c)(c-a) }
&\tiny=\frac { { a }^{ 2 }\{ { b }^{ 3 }-{ c }^{ 3 }-3bc(b-c)\} +{ b }^{ 2 }\{ { c }^{ 3 }-{ a }^{ 3 }-3ca(c-a)\} +c^{ 2 }\{ a^{ 3 }-b^{ 3 }-3ab(a-b)\}  }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-3{ a }^{ 2 }bc(b-c)+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })-3ab^{ 2 }c(c-a)+c^{ 2 }(a^{ 3 }-b^{ 3 })-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }^{ 2 }bc(b-c)-3ab^{ 2 }c(c-a)-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }bc\{ (b-c)+(c-a)+(a-b)\}  }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 }) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b^{ 2 }-c^{ 2 })+{ a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-b^{ 2 }c^{ 2 }(b-c) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b+c)+{ a }^{ 2 }({ b }^{ 2 }+{ bc+c }^{ 2 })-b^{ 2 }c^{ 2 } }{ (a-b)(c-a) }
\end{align}
Would it be possible to solve this answer easily without direct polynomial division? By the use of some known identities, perhaps?
 A: Set $f(a,b,c) = a^2(b-c)^3+b^2(c-a)^3+c^2(a-b)^3$. Note that $f(a,b,c)$ is cyclic, i.e., $f(a,b,c) = f(c,a,b) = f(b,c,a)$.
Further, we also have $$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$$
This means $f(a,b,c) = (a-b)(b-c)(c-a)g(a,b,c)$, where $g(a,b,c)$ is also cyclic polynomial of degree $2$.
The most general cyclic polynomial of degree $2$ in $3$ variables is $$g(a,b,c) = k_1\left(a^2+b^2+c^2\right) + k_2\left(ab+bc+ca\right)$$ Setting $c=0$, we obtain
$$f(a,b,0) = a^2b^3-a^3b^2 = a^2b^2(b-a) = (b-a)abg(a,b,0)$$
This gives us $g(a,b,0) = ab = k_1(a^2+b^2)+k_2ab \implies k_1 =0\text{ and }k_2=1$. Hence, we obtain $g(a,b,c) = ab+bc+ca$.
A: If you don't feel like being clever, brute forcing will also work: the numerator
\begin{align*}
&=a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3\\
&=a^2[(b-a)+(a-c)]^3+b^2(c-a)^3+c^2(a-b)^3\\
&=a^2(b-a)^3+3a^2(b-a)(a-c)(b-c)+a^2(a-c)^3+b^2(c-a)^3+c^2(a-b)^3\\
&=-(c-a)^3(a-b)(a+b)+(a-b)^3(c-a)(c+a)+3a^2(b-a)(a-c)(b-c)
\end{align*}
which is $(a-b)(b-c)(c-a)$ multiplied with
\begin{align*}
&\quad-\frac{(c-a)^2(a+b)}{b-c}+\frac{(a-b)^2(c+a)}{b-c}+3a^2\\
&=\left[-\frac{(c-a)^2a}{b-c}+\frac{(a-b)^2a}{b-c}\right]+\left[-\frac{(c-a)^2b}{b-c}+\frac{(a-b)^2c}{b-c}\right]+3a^2\\
&=a(-2a+b+c)+(-a^2+bc)+3a^2=ab+ac+bc.
\end{align*}
The main thing is that you break the problem into chewable sizes.
