If `a=Sigma_(n=0)^(oo) (x^(3x))/(3n)!,b=Sigma_(n=1)^(oo)(x^(3n-2))/(3n-2!) and C= Sigma_(n=1)^(oo)(x^(3n-1))/(3n-1!)`
then the value of `a^(3)+b^(3)+C^(3)-3abc` is

Options

A. 1
B. 0
C. -1
D. -2

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) where:

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

### Step 1: Find \( a + b + c \)

We start by adding \( a \), \( b \), and \( c \):

\[ a + b + c = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} + \sum_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!} + \sum_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!} \]

Notice that the terms in \( b \) and \( c \) can be adjusted to start from \( n=0 \):

\[ = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} + \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)!} + \sum_{n=0}^{\infty} \frac{x^{3n+2}}{(3n+2)!} \]

This can be combined into a single series:

\[ = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]

### Step 2: Find \( a + b\omega + c\omega^2 \)

Let \( \omega \) be a primitive cube root of unity, i.e., \( \omega = e^{2\pi i / 3} \). We calculate:

\[ a + b\omega + c\omega^2 = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} + \sum_{n=1}^{\infty} \frac{(x\omega)^{3n-2}}{(3n-2)!} + \sum_{n=1}^{\infty} \frac{(x\omega^2)^{3n-1}}{(3n-1)!} \]

This can be rewritten as:

\[ = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} + \sum_{n=0}^{\infty} \frac{(x\omega)^{3n}}{(3n)!} + \sum_{n=0}^{\infty} \frac{(x\omega^2)^{3n}}{(3n)!} \]

Factoring out the common term:

\[ = \sum_{n=0}^{\infty} \frac{1}{(3n)!} \left( x^{3n} + (x\omega)^{3n} + (x\omega^2)^{3n} \right) \]

This simplifies to:

\[ = \sum_{n=0}^{\infty} \frac{(x + x\omega + x\omega^2)^{3n}}{(3n)!} \]

Using the property \( 1 + \omega + \omega^2 = 0 \):

\[ = e^{x(1 + \omega + \omega^2)} = e^0 = 1 \]

### Step 3: Find \( a + b\omega^2 + c\omega \)

By symmetry, we find:

\[ a + b\omega^2 + c\omega = 1 \]

### Step 4: Calculate \( a^3 + b^3 + c^3 - 3abc \)

Using the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \]

Substituting the values we found:

\[ = e^x \cdot 1 \cdot 1 = e^x \]

### Step 5: Evaluate at \( x = 0 \)

To find the final answer, we evaluate at \( x = 0 \):

\[ e^0 = 1 \]

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is:

\[ \boxed{1} \]