Answer
$\Sigma_{n=0}^\infty (-1)^n\dfrac{(x)^{(6n)}}{5^{(n)}(2n)!}$
Work Step by Step
Consider the Taylor Series for $\cos x$ is defined as:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-...$
Replace $(\dfrac{x^3}{\sqrt 5})$ in the above series.
we have $ \cos (\dfrac{x^3}{\sqrt 5})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x^3}{\sqrt 5})^{2n}}{(2n)!}$
This implies that $1-\dfrac{(\dfrac{x^3}{\sqrt 5})^2}{2!}+\dfrac{(\dfrac{x^3}{\sqrt 5})^4}{4!}-...=\Sigma_{n=0}^\infty (-1)^n\dfrac{(x)^{(6n)}}{5^{(n)}(2n)!}$
