I liked this question so much I made it the subject of my blog on June 4th, 2013. Thanks for the great question!
Large cases are easy to come by. For example:
a = 1073741823;
b = 134217727;
c = 134217727;
because b * c overflows to a negative number.
I would add to that the fact that in checked arithmetic, the difference between a / (b * c) and (a / b) / c can be the difference between a program that works and a program that crashes. If the product of b and c overflows the bounds of an integer then the former will crash in a checked context.
For small positive integers, say, small enough to fit into a short, the identity should be maintained.
Timothy Shields just posted a proof; I present here an alternative proof. Assume all the numbers here are non-negative integers and none of the operations overflow.
Integer division of x / y finds the value q such that q * y + r == x, where 0 <= r < y.
So the division a / (b * c) finds the value q1 such that
q1 * b * c + r1 == a
where 0 <= r1 < b * c
the division ( a / b ) / c first finds the value qt such that
qt * b + r3 == a
and then finds the value q2 such that
q2 * c + r2 == qt
So substitute that in for qt and we get:
q2 * b * c + b * r2 + r3 == a
where 0 <= r2 < c and 0 <= r3 < b.
Two things equal to the same are equal to each other, so we have
q1 * b * c + r1 == q2 * b * c + b * r2 + r3
Suppose q1 == q2 + x for some integer x. Substitute that in and solve for x:
q2 * b * c + x * b * c + r1 = q2 * b * c + b * r2 + r3
x = (b * r2 + r3 - r1) / (b * c)
where
0 <= r1 < b * c
0 <= r2 < c
0 <= r3 < b
Can x be greater than zero? No. We have the inequalities:
b * r2 + r3 - r1 <= b * r2 + r3 <= b * (c - 1) + r3 < b * (c - 1) + b == b * c
So the numerator of that fraction is always smaller than b * c, so x cannot be greater than zero.
Can x be less than zero? No, by similar argument, left to the reader.
Therefore integer x is zero, and therefore q1 == q2.