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.
 A: There are a continuum of solutions to
$$
a^3+b^3+3ab=1
$$
Suppose that
$$
x=a+b
$$
then
$$
\begin{align}
1
&=a^3+(x-a)^3+3a(x-a)\\
&=x^3-3ax^2+3a^2x+3ax-3a^2
\end{align}
$$
which means
$$
\begin{align}
0
&=(x-1)\left(x^2+(1-3a)x+3a^2+1\right)
\end{align}
$$
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: 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 $a+b=-2$.
A: 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. 
A: HINT: 
$$(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+3ab(a+b)+b^3$$
Or see this question, especially my answer to it.
A: The three solutions of the equation $a^3+b^3+3ab=1$ are 
$$b_1=1-a,$$
$$b_2,b_3 = \frac{1}{2}\left(a-1\pm i\sqrt{3}(1+a)\right).$$
If $a\neq -1$, since $a,b\in\mathbb{R}$, we must have $a+b=a+b_1=1$. If 
$a=-1$, then the imaginary part of $b_2,b_3$ vanishes and we find two solutions for $(a,b)$: $(-1,-1)$ and $(-1,2)$ so $a+b$ is -2 or 1 respectively.
A: $\newcommand{\+}{^{\dagger}}%
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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,

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

A: Hint:
$$a^3+b^3+3ab=(a+b)^3-3ab(a+b)+3ab\ldots\ldots$$
A: As stated, there is no unique answer. 
Let x = a + b, c = 3ab. Then 
a³ + b³ + 3ab(a + b) = (a + b)³, i.e. 
1 = a³ + b³ + 3ab = (a + b)³ - 3ab(a + b - 1), i.e. 
x³ - cx + c - 1 = 0, i.e. 
(x - 1)(x² + x - c + 1) = 0, i.e. 
x = 1 or x = ½ [± √(4c - 3) - 1], if c ≥ ¾.
A: $$a^3+b^3-1+3ab=(a+b)^3-1-3a^2b-3ab^2+3ab=$$
$$=(a+b-1)((a+b)^2+a+b+1)-3ab(a+b-1)=$$
$$=(a+b-1)(a^2+b^2-ab+a+b+1).$$
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 $$a+b=-2.$$
