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?

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.
