A, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.

Notice that

(1 + x)(1 + y) = 1 + x + y + x y

So we can add 1 to both sides of both equations, and we use the property above to get

a + b + a b = 76  ==>  (1 + a)(1 + b) = 77

and

c + d + c d = 54  ==>  (1 + c)(1 + d) = 55

Now, 77 = 7*11 and 55 = 5*11, so we get

a + 1 = 7  ==>  a = 6

b + 1 = 11  ==>  b = 10

(or the other way around, since the given relations are symmetric)

and

c + 1 = 5  ==>  c = 4

d + 1 = 11  ==>  d = 10

Now substitute these values into the desired quantity:

(a + b + c + d) a b c d = 72,000