Question

If $a=\sum _{n==0}^{\infty }\frac{x^{3n}}{3n!}$, $b=\sum _{n=1}^{\infty }\frac{x^{(3n-2)}}{(3n-2)!}$ and $c=\sum _{n=1}^{\infty }\frac{x^{(3n-1)}}{(3n-1)!}$.Then the value of $a^3+b^3+c^3-3abc$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is A

$1$

The explanation of the correct option :

Step1. Find the value of the value of $\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}$ :

Given $a=\sum _{n==0}^{\infty }\frac{x^{3n}}{3n!}$ ,

$b=\sum _{n=1}^{\infty }\frac{x^{(3n-2)}}{(3n-2)!}$, and

$c=\sum _{n=1}^{\infty }\frac{x^{(3n-1)}}{(3n-1)!}$

Adding, we get

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

$$\begin{gathered} =1+\frac{x^3}{3!}+\frac{x^6}{6!}+\dots .+\frac{x}{1!}+\frac{x^4}{4!}+\dots ..+\frac{x^2}{2!}+\frac{x^5}{5!}+\dots . \\ =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\dots . \\ =e^x \end{gathered}$$

Step2. Find the value of the value of ${\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{+}{\mathit{c}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathit{a}\mathit{b}\mathit{c}$ :

Since$a+b\omega +c{\omega }^2=1+\omega x+\frac{{\omega }^2{x}^2}{2!}+.....$

$=e^{\omega x}$

So $a+b{\omega }^2+c\omega =e^{{\omega }^{2}x},\omega$ is the imaginary root of unity

Now, $a^3+b^3+c^3–3abc=(a+b+c)(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )$

$=e^x{e}^{\omega x}{e}^{\omega x^2}$

$=e^{x(1+\omega +{\omega }^2)}$

$=e^{x.0}\left[\because 1+\omega +{\omega }^2=0\right]$

$=1$

Hence, the correct option is (A).