Theory of equations

Let

$P(x)=x+x^3+x^9+x^{27}+x^{81}+x^{243}$

When we divide $P(x)$ by $x^2-1$ we get a remainder of degree $1$ $\Longrightarrow r(x)=ax+b$ . This translates to

$P(x)=(x+1)(x-1)q(x)+ax+b$

For some polynomial $q$ . We have that $P(1)=6$ and $P(-1)=-6$ , thus

$\begin{cases} P(\mathbin{\color{#D61F06}1})=(\mathbin{\color{#D61F06}1}+1)(\mathbin{\color{#D61F06}1}-1)q(\mathbin{\color{#D61F06}1})+a\cdot\mathbin{\color{#D61F06}1} +b\\ P(\mathbin{\color{#3D99F6}-1})=(\mathbin{\color{#3D99F6}-1}+1)(\mathbin{\color{#3D99F6}-1}-1)q(\mathbin{\color{#3D99F6}-1})+a(\mathbin{\color{#3D99F6}-1}) +b \end{cases}$

$\Longrightarrow\quad \begin{cases} a+b=6\\ -a+b=-6 \end{cases}$

$\Longrightarrow a=6,b=0$ . So the remainder is $\boxed{6x}$