abc = a+b+c³
100a+10b+c=a+b+c³
99a+9b=c³-c
9(11a+b)=c(c²-1)
9(11a+b)=c(c-1)(c+1)
=> c-1=9, sau c=9; sau c+1=9
I. c-1=9 =>c=10; c+1=11
9(11a+b)=9•10•11 /:9
11a+b =110, nu convine, 11a+b ≤108; pentru a=b=9
II. c=9=>c-1=8; c+1=10
9(11a+b)=8•9•10 /:9
11a+b=80 => a=7 si b=3 => abc=739
739=7+3+9³ (A)
III. c+1=9=>c=8; c-1=7
9(11a+b)=7•8•9 /:9
11a+b=56 => a=5 si b=1 => abc=518
518=5+1+8³ (A)
abc∈{518; 739}