How to prove that $3ab(a+b)$ cannot be a cube?

Question

Consider the diophantine equation: $ 3ab(a+b)=c^3 $ where $a,b,c$ are non-zero integers,how do you prove that this equation has no integral solutions?

Answer 1

Edition/Revision 2.

Equation $$ 3ab(a+b) = c^3 $$ has non-zero integer solutions.

One of "obvious" solutions is $$ (a,b,c) = (1,8,6);\tag{A} $$ other (not so obvious) sulutions that I've found so far, are: $$ (a,b,c) = (3087, 4913,7140) \\ \color{gray}{ = (9 \cdot 7^3, 17^3, 3 \cdot 7\cdot 17 \cdot 20)},\tag{B} $$ $$ (a,b,c) = (756249048, 19902511, 327250386) \\ \color{gray}{ = (9 \cdot 438^3, 271^3, 3 \cdot 438 \cdot 271 \cdot 919)},\tag{C} $$ $$ (a,b,c) = (6646883738818239, 48707103808000, 1866552387462840) \\ \color{gray}{ = (9 \cdot 90391^3, 36520^3, 3 \cdot 90391 \cdot 36520 \cdot 188479)}.\tag{D} $$

These solutions can be represented as sum of $3$ cubes (as metacompactness noted in comments to the question): $$ \color{gray}{a^3 + b^3 + c^3 = (a+b)^3;}\\ \color{gray}{1^3 + 8^3 + 6^3 = (1+8)^3 = 9^3;}\\ \color{gray}{3087^3 + 4913^3 + 7140^3 = (3087+4913)^3 = 8000^3;}\\ ... $$

---

Here I consider only solutions with co-prime $a,b$. (It is clear why).

---

Way to find solutions:

we'll focus on co-prime $a,b$.

If $3| c^3$, then $27| c^3$, then $9| ab(a+b)$.

$3$ cases are here:
A) $9\mid a, \quad 3\nmid b$;
A') $3\nmid a, \quad 9\mid b$; (symmetric to case A) )
B) $3\nmid a, \quad 3\nmid b, \quad 9\mid (a+b)$.

case A):
$a,b,(a+b)$ are co-prime pairwice.
So, $a,b,(a+b)$ have different prime factors. Then every of numbers $3a, b, (a+b)$ has each prime factor in $3\times$ power. Other words, $3a,b,(a+b)$ are co-prime cubes:

$$3a = 27 p^3,\qquad b = q^3, \qquad (a+b) = r^3.$$

Then $$3ab(a+b) = 27p^3q^3r^3 = (3pqr)^3.$$ $$c = 3pqr.$$ Thus for searching $(a,b,c)$ one needs to consider co-prime pairs $(p,q)$, such that $$ 9p^3+q^3 = r^3.\tag{*} $$

Triple $(p,q,r) = (7,17,20)$ generates solution $(B)$,
triple $(p,q,r) = (438,271,919)$ generates solution $(С)$,
triple $(p,q,r) = (90391,36520,188479)$ generates solution $(D)$.

case B):
Similar thinking.
$a,b,3(a+b)$ are co-prime cubes:

$$a = p^3,\qquad b = q^3, \qquad 3(a+b) = 27r^3. $$

Then $$ab\cdot3(a+b) = 27p^3q^3r^3 = (3pqr)^3.$$

Thus for searching $(a,b,c)$ one needs to consider co-prime pairs $(p,q)$, where $3\nmid p$ and/or $3\nmid q$, such that $$ p^3+q^3 = 9r^3.\tag{**} $$

Triple $(p,q,r)=(1,2,1)$ generates solution $(A)$.