<s>[INST] Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:

Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\ 
x^2 + y^2 &= 5 \\ 
x^3+y^3 &= 8
\end{align}$$

Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$x^3+y^3 = 4(x+y)$
$x^3+y^3 = 8$
Then:
$8 = 4(x+y)$
$x+y = 2$
However if we replace that value in $(x+y)^2$ we have:
$(x+y)^2 = 2^2 =4$
$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$
As you can see $4 \neq 7$, what is happening?
 [/INST] We can also take a geometric interpretation of the problem. The curves corresponding to the equations
$$ xy \  = \  1 \ \ \ , \ \ \ x^2 \ + \ y^2 \   = \  5 \ \ \ , \ \ \ x^3 \ + \ y^3  \  = \  8 $$
all have symmetry about the line $ \ y \ = \ x \ $ (only the first two, however, has symmetry about the origin).  This suggests an approach in which we look at intersections of these curves with lines through the origin $ \ y \ = \ mx \ \ . \ $  (We do not need to consider a "horizontal" or "vertical" line, since the first curve has the coordinates axes as asymptotes.)  If there  are value(s) of $ \ m \ $ which "work" for all three curves, then we have located their mutual intersections.
With this substitution, we have
$$ m·x^2 \  = \  1 \ \ \ , \ \ \ (1 + m^2)·x^2  \   = \  5 \ \ \ , \ \ \ (1 + m^3)·x^3  \  = \  8 \ \ . $$
For the first two curve equations, we obtain
$$  (1 + m^2)·\frac{1}{m}  \ \  = \ \ 5 \ \ \Rightarrow \ \ m^2 - 5m + 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \ . $$
(As $ \ \frac{2}{5 \ +\ \sqrt{21}} \ = \ \frac{2·(5 \ - \ \sqrt{21})}{25 - 21} \ = \ \frac{5 \ - \ \sqrt{21}}{2} \ \ , \ $ we see that these lines are indeed symmetrical about $ \ y \ = \ x \ \ . \   $   These slopes also tell us that the circle and rectangular hyperbola intersect only in the first and third quadrants.)
However, in the third equation, this gives us
$$ (1 + m^3)·\left( \frac{1}{m} \right)^{3/2}  \ \  \stackrel{?}{=} \  \   8  \ \ \Rightarrow \ \ (1 + m^3)   \ \  \stackrel{?}{=} \  \   8· m^{3/2}  \ \  $$
$$ \Rightarrow \ \ \left( \ 1 \ + \ \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^3 \ \right)   \ \  \stackrel{?}{=} \  \   8· \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^{3/2} $$
$$ \Rightarrow \ \ \ 8 \ + \  \ ( \ 5^3 \ \pm \ 3·5^2·\sqrt{21} \ + \ 3·5·21 \ \pm \ 21^{3/2} \ )    \ \  \stackrel{?}{=} \  \   16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$
$$ \Rightarrow \ \ \ 8 \ + \  \ ( \ 440 \ \pm \ 96·\sqrt{21} \  \ )    \ \  \stackrel{?}{=} \  \   16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$
[then, squaring both sides]
$$ \Rightarrow \ \ \ ( \ 448 \   \pm \ 96·\sqrt{21} \  )^2    \ \  \stackrel{?}{=} \  \   512· ( \ 448 \   \pm \ 96·\sqrt{21} \  ) \ \ \Rightarrow \ \ \  448 \   \pm \ 96·\sqrt{21}     \ \  \neq \  \   512 \ \ . $$
So we may conclude that the pairwise intersections of the three curves do not all fall on the same radial lines from the origin, and thus there is no solution to the given system of non-linear equations.
Interestingly, the second and third curves nearly meet tangentially.  The line $ \ y \ = \ x \ $ intersects the circle at $ \ x^2 \ = \ \frac52 \ \Rightarrow \ \left(\sqrt{\frac52} \ , \ \sqrt{\frac52} \ \approx \ 1.5811 \right) \ \ $ and intersects the "cubic curve" at  $ \ x^3 \ = \ 4 $ $ \Rightarrow \ \left(\sqrt[3]{4} \ , \ \sqrt[3]{4} \ \approx \ 1.5874 \right) \ \ , \ $ so the intersections of these two curves (in the first quadrant -- there is a second "diagonally symmetric" pair in the second and fourth quadrants ) lie on lines with reciprocal slopes close to $ \ 1 \ \ . \ $ (Those slopes are approximately $ \ m \ \approx \ 1.197 \ \ , \ \ m \ \approx \ 0.835 \ \ . \ ) $
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