Given $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$, find the value of $(a+b)$

Question

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.

Answer 1

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.

Answer 2

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 $$a+b=1.$$

Answer 3

$(a+b)^3 = a^3+b^3+3a^2b+3ab^2$

$ 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 $u=1$.