<s>[INST] How find the value of the $x+y$ Question:

let $x,y\in \Bbb R $, and such
  $$\begin{cases}
3x^3+4y^3=7\\
4x^4+3y^4=16
\end{cases}$$
Find the $x+y$

This problem is from china some BBS
My idea: since
$$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$
$$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$
then I can't  Continue
 [/INST] Another idea for you to think about. We have:
$$\begin{cases}
3x^3+4y^3=7 \\
4x^4+3y^4=16
\end{cases}$$
Break it as: 
$$\begin{cases}
3x^3+3y^3 + y^3=7 \\
x^4 + 3x^4+3y^4=16
\end{cases}$$
and factor:
$$\begin{cases}
3(x+y)(x^2 - xy + y^2)=7 - y^3\\
3(x+y)(x^3 - x^2y + xy^2 - y^3)=16 - x^4
\end{cases}$$
And so: $$x+y = \frac{1}{3}\frac{7 - y^3}{x^2 - xy + y^2} = \frac{1}{3}\frac{(2-x)(8+4x+2x^2+x^3)}{x^3 - x^2y + xy^2 - y^3}$$
</s>