If a = sumlimits_{n = 0}^infty {frac{{{x^{3n} | vedclass

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(A) We know that the expansion of $e^x$ is $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.Given $a = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$,$b = \sum_{n=1}^{

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(A) We know that the expansion of $e^x$ is $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.Given $a = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$,$b = \sum_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}$,and $c = \sum_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}$.Summing these,$a+b+c = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$.Using the cube roots of unity $\omega$ and $\omega^2$,we have:$a + b\omega + c\omega^2 = \sum_{n=0}^{\infty} \frac{(\omega x)^n}{n!} = e^{\omega x}$.$a + b\omega^2 + c\omega = \sum_{n=0}^{\infty} \frac{(\omega^2 x)^n}{n!} = e^{\omega^2 x}$.Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$:$a^3 + b^3 + c^3 - 3abc = e^x \cdot e^{\omega x} \cdot e^{\omega^2 x} = e^{x(1+\omega+\omega^2)}$.Since $1+\omega+\omega^2 = 0$,we get $e^{x(0)} = e^0 = 1$.

If $a = \sum\limits_{n = 0}^\infty {\frac{{{x^{3n}}}}{{(3n)!}}} ,\,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 = $

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