297. Find the remainder of the division of the polynomial

$$
x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}
$$

a) by $x-1$; b) by $x^{2}-1$.
297. a) First solution. Since for any non-negative integer $k$ the binomial $x^{k}-1$ is divisible by $x-1$, then

$$
\begin{aligned}
& x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=(x-1)+\left(x^{3}-1\right)+ \\
&+\left(x^{9}-1\right)+\left(x^{27}-1\right)+\left(x^{81}-1\right)+\left(x^{243}-1\right)+6
\end{aligned}
$$

gives a remainder of 6 when divided by $x-1$.

Second solution. Let $q(x)$ be the quotient and $r$ the remainder when $x+x^{3}+$ $+x^{9}+x^{27}+x^{81}+x^{243}$ is divided by $x-1$. Then

$$
x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=q(x)(x-1)+r
$$

Substituting $x=1$ in this equation, we get $6=r$.

b) Analogous to the second solution of the previous problem, let $q(x)$ be the quotient when our polynomial is divided by $x^{2}-1$, and $r_{1} x+r_{2}$ be the desired remainder (the remainder from dividing a polynomial by a quadratic trinomial is a binomial of the first degree):

$$
x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=q(x)\left(x^{2}-1\right)+r_{1} x+r_{2}
$$

Substituting $x=1$ and $x=-1$ in the last equation, we get:

$$
6=r_{1}+r_{2}-6=-r_{1}+r_{2}, \text { from which } r_{1}=6, r_{2}=0
$$

Thus, the desired remainder is $6 x$.