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$$a=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}\\b=\sum_{n=1}^\infty\frac{x^{3n-2}}{(3n-2)!}\\c=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$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.