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.

Using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and denoting $(a+b)$ by $A$, we have (from the assumption)
\begin{align*}
&a^3+b^3+3ab=1\\
&(a+b)(a^2-ab+b^2)+3ab=1\\
&A(A^2-3ab)+3ab=1\\
&(A^3-1)-3abA+3ab=0\\
&(A-1)(A^2+A+1)-3ab(A-1)=0\\
&(A-1)(A^2+A+1-3ab)=0,\\
\end{align*}
where we use the identities $a^2+b^2=(a+b)^2-2ab=A^2-2ab$ and $A^3-1=(A-1)(A^2-A+1)$. 
Therefore, $A=1$ or $A^2+A+1-3ab=0$. 
In the latter case, replacing $A$ by $a+b$, we can conclude that the only solution is $a=b=-1$ (getting $a^2+(1-b)a+(b^2+b+1)=0$ and using the determinant for real solutions of quadratic equations), which is left to you. 
