If `a= .^(20)C_(0) + .^(20)C_(3) + .^(20)C_(6) + .^(20)C_(9) + "…..", ``b = .^(20)C_(1) + .^(20)C_(4) + .^(20)C_(7) + "……"' and ``c = .^(20)C_(2) + .^(20)C_(5) + .^(20)C_(8) + "…..",` then
Value of `a^(3) + b^(3) + c^(3) - 3abc` is

Explanation

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

- \( a = \binom{20}{0} + \binom{20}{3} + \binom{20}{6} + \ldots \)
- \( b = \binom{20}{1} + \binom{20}{4} + \binom{20}{7} + \ldots \)
- \( c = \binom{20}{2} + \binom{20}{5} + \binom{20}{8} + \ldots \)

### Step 1: Understand the sums \( a, b, c \)

The sums \( a, b, c \) consist of binomial coefficients taken at intervals of 3. These can be interpreted as the sums of the coefficients of \( (1 + x)^{20} \) evaluated at the cube roots of unity.

### Step 2: Calculate \( a + b + c \)

We know that:

\[ a + b + c = \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \]

This is because the sum of the binomial coefficients for any \( n \) is \( 2^n \).

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

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

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

Using the binomial theorem, we can expand \( (1 + \omega)^{20} \):

\[ (1 + \omega)^{20} = \sum_{k=0}^{20} \binom{20}{k} \omega^k \]

We can separate the terms based on their residues modulo 3:

\[ (1 + \omega)^{20} = a + b\omega + c\omega^2 \]

### Step 4: Calculate \( (1 + \omega)^{20} \)

Now we calculate \( 1 + \omega \):

\[ 1 + \omega = 1 + \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2} \]

The modulus of \( 1 + \omega \) is:

\[ |1 + \omega| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

The argument is:

\[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \frac{\pi}{3} \]

Thus, we have:

\[ (1 + \omega)^{20} = 1^{20} \cdot e^{20 \cdot \frac{\pi}{3} i} = e^{\frac{20\pi}{3} i} \]

### Step 5: Simplify \( e^{\frac{20\pi}{3} i} \)

To simplify \( e^{\frac{20\pi}{3} i} \):

\[ \frac{20\pi}{3} = 6\pi + \frac{2\pi}{3} \quad \text{(since } 6\pi \text{ is a multiple of } 2\pi\text{)} \]

Thus,

\[ e^{\frac{20\pi}{3} i} = e^{\frac{2\pi}{3} i} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = \omega^2 \]

### Step 6: Calculate \( a + b\omega^2 + c\omega \)

Similarly, we can evaluate \( a + b\omega^2 + c\omega \):

\[ (1 + \omega^2)^{20} = e^{\frac{20\pi}{3} i} = \omega \]

### Step 7: Use the identity for cubes

Using the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)((a + b + c)^2 - 3(ab + bc + ca)) \]

We already know \( a + b + c = 2^{20} \).

### Step 8: Final Calculation

Since \( a + b + c = 2^{20} \) and \( a + b\omega + c\omega^2 = \omega^2 \) and \( a + b\omega^2 + c\omega = \omega \), we can substitute these values back into the identity and simplify.

After all calculations, we find:

\[ a^3 + b^3 + c^3 - 3abc = 2^{20} \]

### Final Answer

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

\[ \boxed{2^{20}} \]