I'm talking about exercises like these for example:
$ (a+2b)^3 - (a-2b)^3 $
$(a+b+c)(a+b-c)(a-b+c)(a-b+c)(-a+b+c)$
Of course these can be done the time-consuming and mentally easy way, but are there some general insights or tricks everyone should now regarding these? I know of the binomial theorem, but that still is time-consuming and I don't think it works for examples like the second one.
On
First problem:
$$\begin{align}(a+2b)^3-(a-2b)^3&=(A+B+C+D)-(A-B+C-D)\\ &=2(B+D)\\ &=2(3a^22b +8b^3)\\ &=4b(3a^2+4b^2)\end{align}$$
Second problem:
This looks similar to Heron's formula for the area of a triangle in terms of its semiperimeter. Some useful info here. http://en.wikipedia.org/wiki/Heron%27s_formula
I don't think there is a general approach to finding the most efficient way to solve such quizzes.
In the first place, you need to know all the classical identities by heart and be able to recognize them even in disguise, like $(a+b-c)(a-b-c)=(a-c)^2-b^2$.
Then you can mentally predict how the computation will proceed and make bets: in your first example, you see a difference of two cubes. This is known to factorize as a first degree binomial and a second degree trinomial. The former is easy. The latter will develop as the sum of two squares and a product of sum/difference; there will be 8 terms in total and a bit of simplification. (Mentally, $3a^2+4b^2$.)
You can also try direct evaluation, by developing the cubes: this will also involve 8 terms, of which 4 will cancel out thanks to the alternating signs and the 4 others will merge to 2. (Mentally: $12a^2b+16b^3$). After putting $4b$ in evidence, a sum of squares remains. Both ways are similar in complexity.