2006 SMT/Algebra Problems/Problem 4

Problem

Simplify: $\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$

Solution

Firstly, we notice that the factors in the denominators are almost the same. To make them the same, we merely have to change a couple of signs, to get

\[\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}\] \[=-\frac{a^3}{(a-b)(c-a)}-\frac{b^3}{(a-b)(b-c)}-\frac{c^3}{(c-a)(b-c)}.\]

Now it becomes easier to combine the fractions. We have

\[=\frac{-a^3(b-c)-b^3(c-a)-c^3(a-b)}{(a-b)(b-c)(c-a)}\]

Now we try expanding the numerator.

\[=\frac{-a^3b+a^3c-b^3c+ab^3-ac^3+bc^3}{(a-b)(b-c)(c-a)}\]

Now we can try to factor the numerator. Notice that letting $a=b, b=c,$ or $a=c$ in the numerator causes it to become $0$, so $(a-b)(b-c)(c-a)$ is a factor of the numerator. Therefore, we only need one more degree of $a, b,$ and $c$ in the numerator. We first try $a+b+c$ as the remaining factor in the numerator, and see that it works. Therefore,

\[=\frac{(a-b)(b-c)(c-a)(a+b+c)}{(a-b)(b-c)(c-a)}=\boxed{a+b+c}\]

See Also

2006 SMT/Algebra Problems