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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.

1 Answers 1

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Put $x=y\omega$ in the expansion of $e^x$ to get$$e^{y\omega}=\sum_{n=0}^\infty\frac{y^{3n}}{(3n)!}+\omega\sum_{n=1}^\infty\frac{y^{3n-2}}{(3n-2)!}+\omega^2\sum_{n=1}^\infty\frac{y^{3n-1}}{(3n-1)!}=a+b\omega+c\omega^2$$since $\omega^{3k}=1,\omega^{3k+1}=\omega,\omega^{3k+2}=\omega^2~\forall k\in\Bbb Z$.

Similarly, put $x=y\omega^2$ in the expansion of $e^x$ to get$$e^{y\omega^2}=a+b\omega^2+c\omega$$Finally, recall that $a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=e^{y(1+\omega+\omega^2)}=1$ since $1+\omega+\omega^2=0$.