This week I wondered about diophantine problems that involve the volume of certain cubes and frustums, see the Wikipedia *Frustum.* I wondered if each one of these problems have infinitely many solutions, I would like to know if it is possible to determine if some of these have infinitely many solutions. I don't know if these problems are easily to get or if these are in the literature.

**Problem 1.** For integers $1<a<b$ and integers $1\leq h$ and $1\leq c$ we consider the diophantine equation $$3(a^3+c^3)=h(a^2+ab+b^2).\tag{1}$$

**Example 1.** One has that $31^3+7^3=30134=\frac{26}{3}(31^2+31\cdot 37+37^2)$.

**Problem 2.** For integers $2\leq x< y< z < v $ and integers $1\leq a < b$ and integer $h\geq 1$ we consider the diophantine equation $$3(4x^3+3y^3+2z^3+v^3)=h(a^2+ab+b^2).\tag{2}$$

**Example 2.** One has that the volume of four cubes of side $x=2$, added to the volume of three cubes of side $y=3$, added to the volume of two cubes $z=4$ and a cube of side $v=5$ is equals to the volume of a square frustum of basis $a=12$ and $b=15$ with height $h=2$, that is $$3(4\cdot 2^3+3\cdot 3^3+2\cdot 4^3+5^3)=1098=2(12^2+12\cdot 15+15^2).$$

Question.Do these problems, Problem 1 or Problem 2 (only is required the approach or reasoning for one of these problems), have infinite solutions?Many thanks.

I've written a small program in Pari/GP for each of these equations, showing more solutions. Each one of these codes are a line of codification that you can run in the web Sage Cell Serverchoosing as Language *GP*, this

`for(a=2, 40, for(b=1, 40, for(c=1, 40, for(h=1, 40, if(a<b&&3*(a^3+c^3)==h*(a^2+a*b+b^2),print(c))))))`

and for the second problem you get the output after two minutes

`for(a=1, 15, for(b=1, 15, for(z=1,15,for(v=1,15, for(h=1, 15, for(x=2, 15, for(y=1, 15, if(a<b&&x<y&&y<z&&z<v&&3*(4*x^3+3*y^3+2*z^3+v^3)==h*(a^2+a*b+b^2),print(v)))))))))`