Solution for $x$ and $y$. If $x^6-y^6 = 665$, and $x^3y+xy^3=78$, find $x$ and $y$.

I have tried this question in the following way:
We know that $x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$. By substituting the values we can yield $665=(x-y)(x^5+78x+78y+y^5)$. Further simplification made it complicated.
These are some initial thoughts. I will add more if I have more ideas.

Let $u = xy$, $v=x+y$, $w=x-y$. Note that $$x^6-y^6=vw\cdot(v^2-u)\cdot(w^2+u)$$And $$x^3y+xy^3=u\cdot(v^2-2u)=u\cdot(w^2+2u)$$This gives us $3$ equations with $3$ variables, and all of the equations are at most cubic.

Let us first limit our search for solutions to the reals. Note that the second equation forces $x,y$ to have the same sign, and by @Arthur's comment, $(x,y)$ being a solution implies $(-x,-y)$ is a solution. So, we can limit our search in the reals to positive $x,y$. 

However, note that $x^3y+y^3x$ is strictly increasing on both $x,y$, so if $x_0,y_0$ and $x_1,y_1$ are both solutions to that equation, $x_0,x_1,y_0,y_1&gt;0$, then WLOG $x_1&gt;x_0$, $y_0&gt;y_1$. This is a problem, however, as if $x_0,y_0$ and $x_1,y_1$ are solutions to the first equation, then $x_1&gt;x_0$ implies $y_1&gt;y_0$. But this is a contradiction, implying that there can be at most one set of positive $x,y$ that solves the system. @DavidG.Stork already found this solution to be $(2,3)$.

So, we've shown that the only real solutions are $\color{red}{(2,3)}$ and $\color{red}{(-2,-3)}$. Solving for complex roots is likely to be tedious.
$x^3y + xy^3 = xy(x^2 + y^2) = 78 = 2*3*13$

$2*3*(3^2+2^2) = 78$

This gives is a good idea where to go in the other equation... do they fit?

$x^6 - y^6 = (x + y)(x^2 - xy + y^2)(x-y)(x^2 + xy + y^2)\\
(3+2)(3^2 - 6 + 4)(3-2)(3^2 + 6 + 2^2) = 5*7*19 = 665$

yes.

Likely not the only solution.

$(-2,-3)$ also works.