2006 SMT/Algebra Problems/Problem 8
Problem
Evaluate:
Solution
Since the denominator of the fraction is factorable, we try partial fraction decomposition. Let . Therefore,
. Plugging in
and
, we find that
.
Thus, . Expanding, we find that this is equal to
.
We now see that everything cancels (two s cancel with one
) except for a
and
in the beginning, and a
, and
in the end.
Therefore, our final answer is .