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.

$(a+b)^3 = a^3+b^3+3a^2b+3ab^2$
$
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 $u=1$.
