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.

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$$
\pars{a + b}^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = 1 - 3ab + 3a^{2}b + 3ab^{2}
=
1 + 3ab\pars{-1 + a + b}
$$
$$
x^{3} -3abx + 3ab - 1 = 0\quad\mbox{where}\quad x \equiv a + b
$$

$$
\pars{x - 1}\pars{x^{2} + x + 1} - 3ab\pars{x - 1} = 0
$$
One solution is $\color{#ff0000}{\large x = a + b = 1}$.

When $x \not=1$ $\pars{~a + b \not= 1~}$:
$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
0 = x^{2} + x + \pars{1 - 3ab} = x^{2} + x + 1 - 3a\pars{x - a}
=
x^{2} + \pars{1 - 3a}x + \pars{1 + 3a^{2}} = 0
\tag{1}
$$
This equation discriminant $\Delta$ is given by:
$$
\Delta = \pars{1 - 3a}^{2} - 4\pars{1 + 3a^{2}}
=
-3a^{2} - 6a - 3 = -3\pars{a + 1}^{2} \leq 0
$$
Since $x = \pars{a + b} \in {\mathbb R}$, this analysis doesn't yield any other solution when $a \not= -1$. When $a = -1$, Eq. $\pars{1}$ has the double root
$x_{\pm} = \pars{3a - 1}/2 = -2$. Then,

$$
\mbox{the solutions are}\quad
{\large\left\lbrace%
\begin{array}{rcl}
a + b & = & \phantom{-}1
\\[1mm]
\mbox{and}\quad a + b & = & -2  \quad\mbox{when}\quad a = -1
\end{array}\right.} 
$$

