Ta có:
\(\left(a^2+b+2\right)a+1=a^3+b^2\\ =>a^3+ab+2a+1=a^3+b^2\\ =>ab+2a+1=b^2\\ =>a\left(b+2\right)-\left(b^2-4\right)=3\\ =>a\left(b+2\right)-\left(b+2\right)\left(b-2\right)=3\\ =>\left(a-b+2\right)\left(b+2\right)=3\)
Vì a,b nguyên nên ta có bảng:
b + 2 | 1 | 3 | -3 | -1 |
a - b + 2 | 3 | 1 | -1 | -3 |
b | -1 | 1 | -5 | -3 |
a | 0 | 0 | -8 | -8 |