Need hint for equation system I stumbled upon the following in an old math book and could use a hint:

The sum of the volumes of two cubes is 35, whereas the sum of the
  surface area of the two cubes is 78. Calculate the lengths of the
  sides of the two cubes.

So, if we let $x$ be the length of the side of one cube, and $y$ be the length of the side of the other, we must have:
$$x^3 + y^3 = 35$$
$$6x^2 + 6y^2 = 78$$
i.e.:
$$x^3 + y^3 = 35$$
$$x^2 + y^2 = 13$$
How to proceed from there? If I try the obvious things like isolating x or y and plug into the other one I get a mess. Is there some type of substitution required?
 A: Hint: $\,(2,3),(3,2)\,$ are "obvious" solutions, then equating $\,x^6 = (35-y^3)^2=(13-y^2)^3\,$ gives:
$$
0 = 2 y^6 - 39 y^4 - 70 y^3 + 507 y^2 - 972 = (y - 2) (y - 3) (2 y^4 + 10 y^3 - y^2 - 135 y - 162)
$$
What's left to prove is that the quartic factor does not have two positive real roots.
A: $35=x^3+y^3=(x+y)(x^2-xy+y^2)=$
$=(x+y)(13-xy)$
But 
$(x+y)^2=x^2+y^2+2xy=13+2xy$ and so
$xy=\frac{1}{2}[(x+y)^2-13]$ and it can be replaced  to first equation to get that 
$35=(x+y)(13-\frac{1}{2}((x+y)^2-13))$
and if you define $z:=x+y$ you must resolve 
$70=z(39-z^2)$
(You can observe that the solutions of the equation are  $z=2$,$z=5$ and $z=-7$)
So the form of the solutions of your equations is $(x,z-x)$ where $z$ is the solution of that particular equation. 
Now you can calulate 
$x^2+(z-x)^2=13$ 
and you have that the unique solutions of your sistem are of the form $(x,z-x)$ such that verify the equation: 
$ 2x^2-2zx+(z^2-13)=0$ 
A: Starting from dxiv's answer
$$2 y^4 + 10 y^3 - y^2 - 135 y - 162=( y^2+7 y+18)(2y^2-4 y-9)$$ The first factor does not show real roots. The roots of the second factor are $y_\pm =1\pm\sqrt{\frac{11}{2}}$.
$y_-$ being discarded since $<0$, computing $x_+$ from $y_+$ would lead to $x_+=1-\sqrt{\frac{11}{2}}<0$. Then the only possible solutions are $(x=2,y=3)$ or $(x=3,y=2)$ .
