I Know the Pieces Fit

Factorizing both expressions, we get the first one is a product and the second one is a sum of $x^3+y^3 = A$ and $x^2+y^2 = B$ .

Because $(A+B)^2 - 4AB = (A-B)^2$ , we have that

$|A-B| = \sqrt{1036^2 - 4 \times 90240}$ $\Rightarrow |A-B| + A + B = 844+1036 \Rightarrow A \vee B = 940$ $A + B - |A-B| = 1036 - 844 \Rightarrow A \vee B = 96$

Supposing $x^3+y^3 = 940$ and $x^2+y^2=96$ , we will have this system:

$(x+y)^2 - 2x = 96$ $(x+y)^3 - 3xy(x+y) = 940$

Isolating $xy$ in the first equation we have $xy = \dfrac{(x+y)^2}{2} - 48$ . Plugging it in the second equation we have

$144(x+y) - \frac{(x+y)^3}{2} = 940$

By trial and error, we figure out $x+y=10$ is a root. This leaves us with also

$\frac{(x+y^2)}{2} + 5(x+y) - 94 = 0$

That yields $(x+y) = \pm \sqrt{213} - 5$ , none of them actually greater than $\boxed{10.}$

PS: For $x^3+y^3 = 96$ and $x^2+y^2=940$ , the values of $x+y$ are not integers.