$$a=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}\\b=\sum_{n=1}^\infty\frac{x^{3n-2}}{(3n-2)!}\\c=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$Find $a^3+b^3+c^3-3abc$:
$(a)\ 1$
$(b)\ 0$
$(c)-1$
$(d)-2$
Please help me solve this question.
I added $a,b$ and $c$. It gives me the expansion of $e^x$.
But i dont know how to use it.