If `a+b+c=6, a^(2) +b^(2)+ c^(2)=16`, find `ab+bc +ca`= ?

Answer

To solve the problem, we need to find the value of \( ab + bc + ca \) given the equations:

1. \( a + b + c = 6 \)
2. \( a^2 + b^2 + c^2 = 16 \)

We can use the identity that relates these quantities:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]

### Step 1: Substitute the known values into the identity

We know \( a + b + c = 6 \), so we can calculate \( (a + b + c)^2 \):

\[ (6)^2 = 36 \]

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

We also know \( a^2 + b^2 + c^2 = 16 \). Now we can substitute these values into the identity:

\[ 36 = 16 + 2(ab + ac + bc) \]

### Step 3: Rearrange the equation to isolate \( ab + ac + bc \)

Subtract 16 from both sides:

\[ 36 - 16 = 2(ab + ac + bc) \]

This simplifies to:

\[ 20 = 2(ab + ac + bc) \]

### Step 4: Solve for \( ab + ac + bc \)

Now, divide both sides by 2:

\[ ab + ac + bc = \frac{20}{2} = 10 \]

Thus, the value of \( ab + ac + bc \) is \( 10 \).

### Final Answer

\[ ab + ac + bc = 10 \]

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