Question

xyz-xy-z cub perfect

Answer 1

Fie $k^{3}=xyz-xy-z=10xy+z-xy-z=9xy= 3^{2} xy= M_{3}$
Rezulta $k^{3}= M_{27}=>xy= M_{3}=3 p^{3};10 \leq xy \leq 99$
De aici p∈(2;3) ⇒ xy∈(24;81) iar z poate fi orice cifra.