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.
There are a continuum of solutions to
$$
a^3+b^3+3ab=1
$$
Suppose that
$$
x=a+b
$$
then
$$
\begin{align}
1
&amp;=a^3+(x-a)^3+3a(x-a)\\
&amp;=x^3-3ax^2+3a^2x+3ax-3a^2
\end{align}
$$
which means
$$
\begin{align}
0
&amp;=(x-1)\left(x^2+(1-3a)x+3a^2+1\right)
\end{align}
$$
So either $x=1$ irregardless of $a$, or
$$
x=\frac{3a-1\pm(a+1)\sqrt{-3}}{2}
$$
Thus, other than $x=1$, the only real $x$ is $-2$, which comes from $a=-1$.

That is, the only two real values of $a+b$ are $1$ and $-2$.
$$a^3+b^3-1+3ab=(a+b)^3-1-3a^2b-3ab^2+3ab=$$
$$=(a+b-1)((a+b)^2+a+b+1)-3ab(a+b-1)=$$
$$=(a+b-1)(a^2+b^2-ab+a+b+1).$$
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 $$a+b=-2.$$