If `a+b+c = 6` and `a^(3) + b^(3) + c^(3) - 3abc = 126`, then `ab + bc + ca` is equal to :

Options

A. 5
B. 6
C. 12
D. 8

Answer

To solve the problem step by step, we will use the given equations and some algebraic identities.

### Step 1: Write down the given equations
We are given:
1. \( a + b + c = 6 \)
2. \( a^3 + b^3 + c^3 - 3abc = 126 \)

### Step 2: Use the identity for the sum of cubes
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Substituting \( a + b + c = 6 \) into the identity gives us:
\[ a^3 + b^3 + c^3 - 3abc = 6(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 3: Set up the equation
From the second given equation, we can substitute:
\[ 126 = 6(a^2 + b^2 + c^2 - ab - ac - bc) \]
Dividing both sides by 6:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 21 \]

### Step 4: Use the square of the sum identity
We know:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting \( a + b + c = 6 \):
\[ 36 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]

### Step 5: Rearranging the equations
From the equation \( a^2 + b^2 + c^2 - ab - ac - bc = 21 \), we can express \( a^2 + b^2 + c^2 \) as:
\[ a^2 + b^2 + c^2 = 21 + ab + ac + bc \]
Now substituting this into the squared sum equation:
\[ 36 = (21 + ab + ac + bc) + 2(ab + ac + bc) \]
This simplifies to:
\[ 36 = 21 + 3(ab + ac + bc) \]

### Step 6: Solve for \( ab + ac + bc \)
Rearranging gives:
\[ 36 - 21 = 3(ab + ac + bc) \]
\[ 15 = 3(ab + ac + bc) \]
Dividing both sides by 3:
\[ ab + ac + bc = 5 \]

### Final Answer
Thus, the value of \( ab + ac + bc \) is \( \boxed{5} \).