compute $a+b$ given $3a^3+4b^3=7$ and $4a^4+3b^4=16$ $a,b\in\mathbb C$, and 
$$3a^3+4b^3=7$$
$$4a^4+3b^4=16$$
compute  $a+b$
I guess there are several $(a,b)$ satisfied the equation set, it is hard to determine all solutions, but $a+b$ should be unique
 A: We can do a straightforward computation to describe the solutions, i.e., without the use of Groebner bases or similar advanced techniques.
Multiply the first equation by $3a/4$ and subtract the second equation to obtain
$$
a=\frac{3(3b^4 - 16)}{4(4b^3 - 7)},
$$
if not $4b^3=7$. But $4b^3=7$ means $a=0$ and $3b^4=16$, a contradiction to $4b^3=7$.
Hence we have $4b^3\neq 7$. 
Then substitute the above expression for $a$ into the two equations. Then we obtain only one polynomial equation in $b$, namely 
$$
18571b^{12} - 114688b^9 - 34992b^8 + 301056b^6 + 186624b^4 - 351232b^3 - 178112=0.
$$
We can write this more nicely as
$$
81(3b^4-16)^3+64(4b^3-7)^4=0.
$$
This has $12$ complex roots (counting with multiplicities), as we know. 
So there are exactly $12$ solutions $(a,b)$, and thus also for $a+b$, which is equal to $\frac{25b^4 - 28b - 48}{4(4b^3 - 7)}$.
We have $2$ real solutions for $b$ and $10$ complex roots. It is easy to see that $a+b$ is not "unique", since we obtain a real value for $b$ real, and a non-real value for the other complex roots of $b$.
