How find the value of the x+yx+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

Answer

Assuming the question is typed correctly as shown, there are two unique real solutions. Let \begin{align*} u(z) &= -983749-111132z^3+786432z^4+71442z^6-196608z^8-20412z^9+18571z^{12}, \\ v(z) &= -178112-351232z^3+186624z^4+301056z^6-34992z^8-114688z^9+18571z^{12}. \end{align*} These polynomials have exactly two distinct real roots; let r(u,+), r(u,-) be the positive and negative real roots of u, and r(v,+), r(v,-) be the positive and negative real roots of v, respectively. Then (x,y) \in \{(r(u,-),r(v,+)), (r(u,+),r(v,-))\} are the desired solutions. The sum x+y can then be expressed by the solution to a third polynomial f(z) = 819447-537600z-8998752z^3+3291428z^3+22132992z^4-17875200z^5+3163146z^6+1042512z^8-437500z^9+18571z^{12}, for which there are again two real roots, both positive. All of these polynomials are irreducible. So I highly doubt that this is a problem that can be reasonably solved by hand.

Attribution
Source : Link , Question Author : math110 , Answer Author : heropup

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