<s>[INST] 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?
 [/INST] 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 $g(a,b,c) = ab+bc+ca$.
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