x³·y=1512-x³ ⇔
x³·y+x³=1512 ⇔
x³(y+1)=1512
Descompun pe 1512 in produs de factori si obtinem
1512=3³·7·2³ ⇔1512=6³·7
Avem urmatoarele cazuri
x³=6³ ⇒x=6; y+1=7 ⇒y=6;
x³=2³ ⇒x=2; y+1=3³·7=27·7=189 ⇒y=188;
x³=3³ ⇒x=3; y+1=2³·7=8·7=56 ⇒y=55;
In concluzie
(x;y)∈{(6;6),(2;188),(3;55)}
