Answer
$1-\dfrac{3}{4}x^{(3)}+\dfrac{9}{16}x^{(6)}-\dfrac{27}{64}x^{(9)}+\dfrac{81}{256}x^{(12)}$
Work Step by Step
Re-write as: $\dfrac{1}{1+\dfrac{3}{4}x^3}=\dfrac{1}{1-(-\dfrac{3}{4}x^3)}$
Consider the Taylor's Series for $\dfrac{1}{1-x}$ as follows:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$
Plug $(-\dfrac{3}{4}x^3)$ instead of $x$ in the above series as follows:
$\dfrac{1}{1-(-\dfrac{3}{4}x^3)}=1+(-\dfrac{3}{4}x^3)+(-\dfrac{3}{4}x^3)^2+(-\dfrac{3}{4}x^3)^3+(-\dfrac{3}{4}x^3)^4+...$
$\implies 1-\dfrac{3}{4}x^{(3)}+\dfrac{9}{16}x^{(6)}-\dfrac{27}{64}x^{(9)}+\dfrac{81}{256}x^{(12)}$
