question_answer

If \[a=\sum\limits_{n=0}^{\infty }{\frac{{{x}^{3n}}}{(3\,\,n)!}},\] \[b=\sum\limits_{n=1}^{\infty }{\frac{{{x}^{3n-2}}}{(3n-2)!}}\] and \[c=\sum\limits_{n=1}^{\infty }{\frac{{{x}^{3n-1}}}{(3n-1)!}},\] then the value of \[{{a}^{3}}+{{b}^{3+}}{{c}^{3}}-3abc\]=

A)  \[1\]    

B)  \[0\]

C)  \[-1\]                           

D)  \[-2\]

E)  None of these