2006 SMT/Algebra Problems/Problem 4
Problem
Simplify: 
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
![\[\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}\]](http://latex.artofproblemsolving.com/6/6/9/6697418c66de001ffdd476ad1dc8d1dc8a9dc71d.png)
![\[=-\frac{a^3}{(a-b)(c-a)}-\frac{b^3}{(a-b)(b-c)}-\frac{c^3}{(c-a)(b-c)}.\]](http://latex.artofproblemsolving.com/3/3/3/333bc26bcc6a7bdc468e4dc6ac5709ce227bc2a8.png)
Now it becomes easier to combine the fractions. We have
![\[=\frac{-a^3(b-c)-b^3(c-a)-c^3(a-b)}{(a-b)(b-c)(c-a)}\]](http://latex.artofproblemsolving.com/2/2/1/221aaf1db773d2f68710602501a750f1e0bd21a8.png)
Now we try expanding the numerator.
![\[=\frac{-a^3b+a^3c-b^3c+ab^3-ac^3+bc^3}{(a-b)(b-c)(c-a)}\]](http://latex.artofproblemsolving.com/e/4/b/e4b3d79fe173393dc1a458e83ababe099d6e5a88.png)
Now we can try to factor the numerator. Notice that letting 
or 
in the numerator causes it to become 
, so 
is a factor of the numerator. Therefore, we only need one more degree of 
and 
in the numerator. We first try 
as the remaining factor in the numerator, and see that it works. Therefore,
![\[=\frac{(a-b)(b-c)(c-a)(a+b+c)}{(a-b)(b-c)(c-a)}=\boxed{a+b+c}\]](http://latex.artofproblemsolving.com/c/6/1/c6181dda5a20ee4118268e0f9418e73880b23cd8.png)
