Calculus

Question

​​If $a=\Sigma_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}, \: b=\Sigma_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!} $ and $c=\Sigma_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!} $, then the value of $(a^3+b^3+c^3-3abc)$ is

• A. 1
• B. 0
• C. -1
• D. -2