If `a+b+c=8` and `ab+bc+ca=20`, then `a^(3)+b^(3)+c^(3)-3abc` is equal to :

Options

A. 30
B. 24
C. 32
D. 36

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given that \( a + b + c = 8 \) and \( ab + bc + ca = 20 \).

### Step 1: Use the identity for \( a^3 + b^3 + c^3 - 3abc \)
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c) \left( (a + b + c)^2 - 3(ab + ac + bc) \right) \]

### Step 2: Substitute the known values
From the problem, we know:
- \( a + b + c = 8 \)
- \( ab + ac + bc = 20 \)

Substituting these values into the identity:
\[ a^3 + b^3 + c^3 - 3abc = 8 \left( 8^2 - 3 \times 20 \right) \]

### Step 3: Calculate \( 8^2 \)
Calculate \( 8^2 \):
\[ 8^2 = 64 \]

### Step 4: Calculate \( 3 \times 20 \)
Calculate \( 3 \times 20 \):
\[ 3 \times 20 = 60 \]

### Step 5: Substitute back into the equation
Now substitute back into the equation:
\[ a^3 + b^3 + c^3 - 3abc = 8 \left( 64 - 60 \right) \]

### Step 6: Simplify the expression
Simplify the expression inside the parentheses:
\[ 64 - 60 = 4 \]

### Step 7: Final calculation
Now calculate:
\[ a^3 + b^3 + c^3 - 3abc = 8 \times 4 = 32 \]

Thus, the final answer is:
\[ \boxed{32} \]