Problem Solving

If 2s = a+b+c then the value of [s(s-a)]/3 + [s(s-b)(s-c)]/3 is

a)ab/3
b)bc/3
c)ca/3
d)abc/3
e)a+b+c/3

It's easiest to pick numbers.

a = 2, b = 3, c = 7

which makes s = 6 and our target 6*4/3 + 6*3*-1/3, or 2.

Plugging into the answers, we find that only A works, so we're done.

EDIT: See my response below. After trying to make a second post with an algebraic explanation, I discovered that none of the answers are correct.

Disturbingly, however, other sets of numbers give conflicting results!

Suppose a = 2, b = 2, c = 2, and s = 3.

Then s*(s - a)/3 + s*(s - b)*(s - c)/3 = 1 + 1 = 2

Either:

1) This isn't a well-formulated question, and you shouldn't use this source;
2) There's a typo in the question stem.