It is not true. Let us call an integer nice if its prime divisors are congruent to $1$ modulo $3$. If $n$ is nice, then in the ring of Eisenstein integers it factors as $n=(c-b\omega)(c-b\bar\omega)$, where the factors $c-b\omega$ and $c-b\bar\omega$ are coprime. In particular, $(b,c)=1$ and we have $n=c^2+cb+b^2$. Note that $(b,c)=1$ implies that $n,b,c$ are pairwise coprime.

Now let $a$ be nice, then $n=a^3$ is also nice, hence by the above there exists a representation $$a^3=c^2+cb+b^2=(c-b)^2+3cb$$ such that $a,b,c$ are pairwise coprime.

It is not true. Let us call an integer nice if its prime divisors are congruent to $1$ modulo $3$. If $n$ is nice, then in the ring of Eisenstein integers it factors as $n=(c-b\omega)(c-b\bar\omega)$, where the factors $c-b\omega$ and $c-b\bar\omega$ are coprime. In particular, $(b,c)=1$ and we have $n=c^2+cb+b^2$. Note that $(b,c)=1$ implies that $n,b,c$ are pairwise coprime.

Now let $a$ be nice, then $n=a^3$ is also nice, hence by the above there exists a representation $$a^3=c^2+cb+b^2=(c-b)^2+3cb$$ such that $a,b,c$ are pairwise coprime.

Here is a concrete counterexample: $a=7$, $b=-18$, $c=19$.