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.

Hint: \begin{align} x^3+y^3+z^3-3xyz& =(x+y+z)(x^2+y^2+z^2-xy-xz-yz) \\ &=(x+y+z)\left(\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}\right) \end{align}
Solution:
$$0=a^3+b^3+(-1)^3-3(a)(b)(-1)=(a+b-1)\left(\frac{(a-b)^2+(a+1)^2+(b+1)^2}{2}\right)$$
so $a+b=1$ or $a=b=-1$. The latter gives $a+b=-2$.
