If a +b + c = 19, ab + bc + ca = 120 then what is the value of `a^3 + b^3 + c^3 - 3abc?`

Options

A. 18
B. 23
C. 31
D. 19

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given the equations \( a + b + c = 19 \) and \( ab + bc + ca = 120 \).

### Step-by-step Solution:

1. **Recall the formula**:
The formula we will use is:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c) \left( (a + b + c)^2 - 3(ab + ac + bc) \right) \]

2. **Substitute the known values**:
From the problem, we know:
- \( a + b + c = 19 \)
- \( ab + ac + bc = 120 \)

Plugging these values into the formula:
\[ a^3 + b^3 + c^3 - 3abc = 19 \left( 19^2 - 3 \times 120 \right) \]

3. **Calculate \( 19^2 \)**:
\[ 19^2 = 361 \]

4. **Calculate \( 3 \times 120 \)**:
\[ 3 \times 120 = 360 \]

5. **Substitute back into the equation**:
Now substitute these values back into the equation:
\[ a^3 + b^3 + c^3 - 3abc = 19 \left( 361 - 360 \right) \]

6. **Simplify the expression**:
\[ 361 - 360 = 1 \]
Therefore,
\[ a^3 + b^3 + c^3 - 3abc = 19 \times 1 = 19 \]

### Final Answer:
Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( 19 \).