Find $a,b,c \in \Bbb N$ such that $n^3+a^3=b^3+c^3$ Is it true that for every $n \in \Bbb N$, there exist  $a,b,c \in \Bbb N$ satisfing
$$
n^3+a^3=b^3+c^3,
$$
where $\operatorname{gcd}(a,b,c)=1$ and $b,c\ne n$?
I checked all positive integers less than 1000, it seems true but I don't know how to prove it.
We have $n^3+(3 n^3 + 3 n^2 + 2 n)^3=(3 n^3 + 3 n^2 + 2 n + 1)^3 - (3 n^2 + 2 n + 1)^3$, but there is a negative number on the right-hand side.

A quick check shows that we have
$$
n^3 + (3 n^3 - 3 n^2 + 2 n)^3 = (3 n^3 - 3 n^2 + 2 n  -1)^3 + (3 n^2 - 
    2 n + 1)^3
$$
Edit 1: If I did not make any mistakes this is the only cubic parameterization. There are no quadratic ones.
For $n\in\mathbb N \implies n\geq 1$, writing $a,b,c$ as
$$
\begin{align}
a &= 3 n^3 - 3 n^2 + 2 n = n (2 n^2 - 1) + (n - 1)^3  + 1\\
b &= 3 n^3 - 3 n^2 + 2 n  -1 = n (2 n^2 - 1) + (n - 1)^3\\
c &= 3 n^2 - 2 n + 1 = 2n^2 + (n-1)^2
\end{align}
$$
we see that $a,b,c \in \mathbb N$ and clearly $c\neq n$. Finally, since
$$
a - b = 1,
$$
we must always have
$$
\gcd(a,b,c) = 1
$$
Therefore the statement is true.

This is found via a parametric search of 
$$
\begin{align}
a &= a_3 n^3 + a_2 n^2 + a_1 n + a_0\\
b &= b_3 n^3 + b_2 n^2 + b_1 n + b_0\\
c &= c_3 n^3 + c_2 n^2 + c_1 n + c_0
\end{align}
$$
In particular the coefficients of $n^9$ and $n^0$ satisfies
$$
\begin{align}
a_3^3 &= b_3^3 + c_3^3\\
a_0^3 &= b_0^3 + c_0^3
\end{align}
$$
so by Fermat's Last Theorem we can assume one of $\{a_3,b_3,c_3\}$ and one of $\{a_0,b_0,c_0\}$ to be $0$ to speed things up a bit.  
