If `a + b + c = 6, a^(2) + b^(2) + c^(2) = 30` and `a^(3) + b^(3) + c^(3) = 165`, then the value of 4abc is:

Options

A. `- 1`
B. `-4`
C. 1
D. 4

Answer

To solve the problem, we will use the relationships between the sums of powers of the variables \(a\), \(b\), and \(c\). We are given:

1. \(a + b + c = 6\)
2. \(a^2 + b^2 + c^2 = 30\)
3. \(a^3 + b^3 + c^3 = 165\)

We need to find the value of \(4abc\).

### Step 1: Use the identity for \(a^3 + b^3 + c^3\)

The identity we will use is:

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

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

First, we calculate \((a + b + c)^2\):

\[ (a + b + c)^2 = 6^2 = 36 \]

### Step 3: Substitute into the identity

Now, substitute the known values into the identity:

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

Substituting the values we have:

\[ 165 - 3abc = 6 \left( 30 - 36 \right) \]

### Step 4: Simplify the equation

Now simplify the right side:

\[ 165 - 3abc = 6 \times (-6) = -36 \]

### Step 5: Solve for \(abc\)

Now rearranging gives:

\[ 165 + 36 = 3abc \]
\[ 201 = 3abc \]
\[ abc = \frac{201}{3} = 67 \]

### Step 6: Find \(4abc\)

Now we can find \(4abc\):

\[ 4abc = 4 \times 67 = 268 \]

### Final Answer

Thus, the value of \(4abc\) is:

\[ \boxed{268} \]