<s>[INST] If $a^3 + b^3 +3ab = 1$, find $a+b$ 
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.

I tried to factorize it but unable to do it.
 [/INST] $$a^3+b^3-1+3ab=(a+b)^3-1-3a^2b-3ab^2+3ab=$$
$$=(a+b-1)((a+b)^2+a+b+1)-3ab(a+b-1)=$$
$$=(a+b-1)(a^2+b^2-ab+a+b+1).$$
Thus, $$a+b=1$$ or
$$a^2+b^2-ab+a+b+1=0$$ or
$$(a-b)^2+(a+1)^2+(b+1)^2=0,$$ which gives $a=b=-1$ and $$a+b=-2.$$
</s>