D.C.B. Marsh proposed this problem in the American Mathematical Monthly in 1957:

“Solve a3b3c3 = 3abc, a2 = 2(b + c) simultaneously in positive integers.”

There are a number of ways to go about this, but Raymond Huck of Marietta College found a strikingly simple one. 3abc is positive, so the first equation tells us immediately that b < a and c < a. Add these two facts together and we get b + c < 2a, and hence 2(b + c) < 4a. Substituting this conveniently into the second equation, we learn that a2 < 4a and a < 4. The second equation also shows that a is an even number, so a must be 2, and b and c, which are smaller, must both be 1.

(“Solutions,” American Mathematical Monthly 65:1 [January 1958], 43-46, Problem E1266, via Ross Honsberger, Mathematical Morsels, 1979.)