Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$? I have tried many times to evaluate $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ as polynomial for some values of integer $n$ which are greater than $1$ for the solution of the titled equation for looking why exactly $c$ always closed to integer which is $-2*10^n$ looking to the behavior of polynomial coefficients but I can't expand that, For numerical evidence I have used such that an example for $n=2$ is  montioned here, Now my question here is :
Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ ? And does this gives any evidence about unknown representation numbers as sum of cubic like  $n=390,732,\cdots $, because What I have tried the decimal expansion of $c$ increases with the digits $9$ ?
Addendum:This part is the motivation of this question, Now I should deduce this question for Unknown representations numbers as $390$, Could we have an integer $c$ for $n=390$ in the titled equation since we have increasing  decimal expansion of $c$ with $999999999\cdots$?
 A: Your expression is
$$(1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 = n \tag{1}\label{eq1A}$$
for $n \gt 1$. For some real $d$, let
$$c = d \times 10^n \tag{2}\label{eq2A}$$
Note that within each of the $3$ terms on the left side of \eqref{eq1A}, there's a term of $1$ or $-1$, but the other terms are all at least $c^2$. If $d$ in \eqref{eq2A} is near $0$, this gives you other solutions, such as the $2$ that Wolfram Alpha gives of $c \approx \pm 0.134$ in your link.
Assume $d$ is not very close to $0$ from this point on (such as $|d| \gt 1$), so you have $c^2 \gg 1$. To make the analysis easier, for now ignore those terms of $\pm 1$. As such, using this and \eqref{eq2A} in the LHS of \eqref{eq1A}, gives
$$\begin{equation}\begin{aligned}
& (1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 \\
& \approx (c^2)^3 + ((c^2)(c + 10^n))^3 + (10^n c^2)^3 \\
& = ((d\times10^n)^2)^3 + (((d\times10^n)^2)(d\times10^n + 10^n))^3 + (10^n (d\times10^n)^2)^3 \\
& = d^6\times 10^{6n} + d^6(d+1)^3\times 10^{9n} + d^6\times10^{9n} \\
& = d^6\times 10^{6n}(1 + 10^{3n}((d+1)^3 + 1))
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Next, you have $d^6\times 10^{6n} \gg n$, so the value within the outside brackets in \eqref{eq3A} should be close to $0$. Also, for $n \gt 1$, you have $10^{3n} \gg 1$, so this means you will then need to have
$$\begin{equation}\begin{aligned}
10^{3n}((d+1)^3 + 1) & \approx 0 \\
(d + 1)^3 + 1 & \approx 0 \\
d + 1 & \approx -1 \\
d & \approx -2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
With $d = -2$, so $c = -2 \times 10^n$, note in the LHS of \eqref{eq1A} that the first term is negative, but much less in magnitude than either of the next two terms. Also, the second term is negative, with it being just slightly less than $-64 \times 10^{9n}$. Finally, the third terms is positive, just about equal in magnitude to the second term, but just slightly less than $64 \times 10^{9n}$. Thus, the overall result is negative. However, as indicated in \eqref{eq3A} and \eqref{eq4A}, as $d$ increases, the overall result increases, so you would expect a value just a bit greater than $c = -2 \times 10^n$ to be a root of \eqref{eq1A}. In fact, in your Wolfram Alpha result, if you click on the "More Digits" button for the $c = -200$ result, you get that $c \approx -199.99991666909716587$.
