72. Find two numbers such that their product, added to each of the given numbers, forms the cube of some number.
72. Diophantus's Solution. Let the first number be $8x$. Then, taking the second to be $x^{2}-1$, we will satisfy one of the requirements, since the product of these numbers $8x^{3}-8x$, added to $8x$, will make $8x^{3}$, i.e., the cube of some number. It is necessary that this same product, added to the second number, should also form the cube of some number, i.e., that $8x^{3}+x^{2}-8x-1$ should be a cube. Assuming that this expression equals $(2x-1)^{3}=8x^{3}-12x^{2}+6x-1$, and solving the equation, we find:

$$
x=\frac{14}{13}
$$

Therefore:

$$
\text { the first number } 8 \cdot \frac{14}{13}=\frac{112}{13}, \text { and the second }\left(\frac{14}{13}\right)^{2}-1=\frac{27}{169}
$$

In the second condition, Diophantus chooses the number so that in the resulting cubic equation, the terms with the cubes of the unknown are eliminated in both parts of the equation.