If `a^(3)+b^(3)+c^(3)-3abc=126 , a+b+c=6` then the value of `(ab+bc+ca)` is :

Options

A. 8
B. 7
C. 5
D. 9

Answer

To solve the problem, we will use the given equations and manipulate them step by step.

### Given:
1. \( a^3 + b^3 + c^3 - 3abc = 126 \)
2. \( a + b + c = 6 \)

### Step 1: Use the identity for the sum of cubes
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Let \( s = a + b + c \) and \( p = ab + ac + bc \). Then we can rewrite the equation as:
\[ a^3 + b^3 + c^3 - 3abc = s \left( a^2 + b^2 + c^2 - p \right) \]

### Step 2: Substitute the known values
From the problem, we know:
\[ s = 6 \]
Thus, we can substitute \( s \) into the equation:
\[ 126 = 6 \left( a^2 + b^2 + c^2 - p \right) \]

### Step 3: Simplify the equation
Dividing both sides by 6 gives:
\[ a^2 + b^2 + c^2 - p = 21 \]
This can be rearranged to:
\[ a^2 + b^2 + c^2 = 21 + p \tag{1} \]

### Step 4: Use the square of the sum
We know that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting \( s = 6 \):
\[ 6^2 = a^2 + b^2 + c^2 + 2p \]
This simplifies to:
\[ 36 = a^2 + b^2 + c^2 + 2p \tag{2} \]

### Step 5: Substitute equation (1) into equation (2)
Now we substitute \( a^2 + b^2 + c^2 \) from equation (1) into equation (2):
\[ 36 = (21 + p) + 2p \]
This simplifies to:
\[ 36 = 21 + 3p \]

### Step 6: Solve for \( p \)
Subtract 21 from both sides:
\[ 15 = 3p \]
Dividing both sides by 3 gives:
\[ p = 5 \]

### Conclusion
Thus, the value of \( ab + ac + bc \) is \( 5 \).