Đáp án:
$M=4.2018$
Giải thích các bước giải:
$a^3+b^3+3(a^2+b^2)+4(a+b)+4=0$
$\rightarrow (a^3+3a^2+3a+1)+(b^3+3b^2+3b+1)+(a+b)+2=0$
$\rightarrow (a+1)^3+(b+1)^3+(a+b)+2=0$
$\rightarrow (a+1+b+1)((a+1)^2-(a+1)(b+1)+(b+1)^2)+(a+b+2)=0$
$\rightarrow (a+b+2)((a+1)^2-(a+1)(b+1)+(b+1)^2+1)=0$
$\rightarrow a+b+2=0$
Do
$(a+1)^2-(a+1)(b+1)+(b+1)^2+1=(a+1-\dfrac{b+1}{2})^2+\dfrac{3(b+1)^2}{4}+1>0\quad\forall a,b$
$\rightarrow a+b=-2$
$\rightarrow M=2018.(-2)^2=4.2018$
