Maclaurin series of $(1+x^3)/(1+x^2)$

The answer you are conjecturing can be written as

$$\frac{1+x^3}{1+x^2}=1-x^2+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})$$

Multiplying both sides by $1+x^2$ gives us

\begin{align*} 1+x^3&=(1+x^2)(1-x^2)+(1+x^2)\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})\\ &=1-x^4+\sum_{n=0}^\infty (-1)^n(1+x^2)(x^{2n+3}+x^{2n+4})\\ &=1-x^4+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6}) \end{align*}

The sum on the RHS is telescoping; that is, $\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})$ is equal to

\begin{align*} (x^3+x^4+x^5+x^6)-(x^5+x^6+x^7+x^8)+(x^7+x^8+x^9+x^{10})-\dots=x^3+x^4. \end{align*}

Therefore,

$$1+x^3=1-x^4+x^3+x^4=1+x^3.$$

You should be able to work backwards to fomulate a proof.

$$\frac{1+x^3}{1+x^2} =x+\frac{1-x}{1+x^2}=x+\frac{1}{1+x^2}-\frac{x}{1+x^2}$$ So you can use the Maclaurin series for $\frac{1}{1+x^2}$ to get the answer.