Mathematics and OrigamiIf x is the side of the smaller cube, the imposed condition will be expressed:
x^3 +()( )( )x+ 13 + x+ 23 = x+ 33
developing and simplifying this expression, we get the following third degree
incomplete equation:
x^3 − 6 x− 9 = 0
We can resolve this equation in three different ways:- Through conventional algebraic means.
- By paperfolding (see Point 7.11 for a resultant values of a = 1, b = 0; c = -6;
d = -9). - Applying a diophantine criterion.
The latter consists in outlining the condition, implicit in the original approach, that the
solution must be a positive integer. As the smaller number of this kind is 1, we can produce a
table with the successive values of x from 1 on, relating them to those taken by both members
of equation x^3 = 6 x+ 9 which is the same established before.
xx^3 6x + 9 ∆1 1 15 -14
2 8 21 -13
32727 0
46433 31
5 125 39 86
6 216 45 171
......................................................
We can also see in the table how the difference between both members of the equation
evolves. From x = 1 up to 3 the difference decreases to zero (which denotes that x = 3 is the
solution), whereas from x = 3 on, the difference increases: this means that there are no more
solutions. The reader can check that 3 is also the result obtained by folding according to Point
7.11.
The exercise will consist then in the construction of small cubes (Point 18.8.1) of side 1
up to a quantity of 432 (teamwork). With half of the total we shall construct cube 6 and with
the other half, cubes 3, 4 and 5. Once the four big cubes 3, 4, 5 and 6 have being built by using
all the small ones, we can verify that the original condition has being fulfilled:
33 + 43 + 53 = 63 ; 27 + 64 + 125 = 216