Given $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$, find the value of $(a+b)$ 
Given: $(a,b)\subset \mathbb R^2$, $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$.
Find: The value of $(a+b)$

Question on the Brazilian Math Olympic (OBM), level 2, phase 3, 2012. No answer provided.
By inspection I can easily see that $(a,b)=(1,0)$, $(a,b)=(0,1)$, are possible solutions, leading to $(a+b)=1$. But are there other solutions? I developed both terms using usual identities, but I'm missing something.
Hints and solutions are appreciated. Sorry if this is a duplicate.
 A: Let $a+b=2u$ and $ab=v^2$.
Thus, $$2u(4u^2-3v^2)=1$$ and
$$2u(v^2+2u+1)=2.$$
From the last we obtain $$v^2=\frac{1}{u}-2u-1$$
Id est,
$$2u\left(4u^2-3\left(\frac{1}{u}-2u-1\right)\right)=1$$ or
$$(2u-1)(4u^2+8u+7)=0$$ or
$$2u=1$$ or
$$a+b=1.$$
A: $(a+b)^3 = a^3+b^3+3a^2b+3ab^2$
$
2=(a+b)(a+1)(b+1)
=(a^2 b + a b^2) + (a+b)^2+ (a+b)
$
Let $u=a+b$ and $v=a^2 b + a b^2$.
Then $u^3=1+3v, 2=v+u^2+u$ and so
$u^3 + 3 u^2 + 3 u - 7=0$.
Now, $u^3 + 3 u^2 + 3 u - 7 = (u+1)^3-8$. Therefore, $u+1=2$ and $u=1$.
A: The inherent symmetries suggest to introduce the elementary symmetric functions of $a$ and $b$ as new variables:
$$a+b=:u,\qquad ab=:v\ .$$
Then
$$a^3+b^3=(a+b)(a^2-ab+b^2)=u(u^2-3v),\qquad(a+1)(b+1)=v+u+1\ .$$
The two given equations then amount to
$$u(u^2-3v)=1,\qquad u(v+u+1)=2\ .\tag{1}$$
In particular $u\ne0$, hence we can solve the second equation $(1)$ for $v$ and obtain
$$v={2\over u}-u-1\ .$$
Plugging this into the first equation $(1)$ we obtain
$$u\left(u^2-3\left({2\over u}-u-1\right)\right)=1\ ,$$
or
$$(u+1)^3=8\ .$$
The only real solution of this equation is $u=1$, and this is also the only possible real value of $a+b$. In order to make sure that real $a$, $b$ satisfying the original conditions indeed exist note that $u=1$ implies $v=0$, and it is then easy to check that $(a,b)\in\{(0,1),(1,0)\}$ are o.k.
