Prove that below equation has only one solution in integers, (0, 0, 0).

Question

Prove that $9a^3 + 3b^3 + c^3 = 0$ has only one solution in integers, $(0, 0, 0)$.

Answer 1

Working modulo 3, first we derive that $3\mid c^3,$ and so $3\mid c$ as 3 is a prime. Then working modulo 9 gives us $9\mid 3b^3,$ and so $3\mid b.$ Finally, considering modulo 27, one gets $27\mid 9a^3\Rightarrow3\mid a.$ Thus, for every triple of solution $(a,b,c)$, there is another, smaller triple of solution $\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right),$ and as the your tag suggests, this leads to an infinite descent and so the only integral solution is $(0,0,0).$