If `a-b-c= 3 and a^(2) + b^(2) + c^(2) = 77`, find: `ab- bc+ ca`

Answer

To solve the problem, we are given two equations:

1. \( a - b - c = 3 \) (Equation 1)
2. \( a^2 + b^2 + c^2 = 77 \) (Equation 2)

We need to find the value of \( ab - bc + ca \).

### Step 1: Rewrite Equation 1
From Equation 1, we can express \( a \) in terms of \( b \) and \( c \):
\[ a = b + c + 3 \]

### Step 2: Substitute \( a \) in Equation 2
Now, substitute \( a \) in Equation 2:
\[ (b + c + 3)^2 + b^2 + c^2 = 77 \]

### Step 3: Expand the left-hand side
Expanding \( (b + c + 3)^2 \):
\[ (b + c + 3)^2 = b^2 + c^2 + 6b + 6c + 9 \]
So, we have:
\[ b^2 + c^2 + 6b + 6c + 9 + b^2 + c^2 = 77 \]
This simplifies to:
\[ 2b^2 + 2c^2 + 6b + 6c + 9 = 77 \]

### Step 4: Simplify the equation
Subtract 9 from both sides:
\[ 2b^2 + 2c^2 + 6b + 6c = 68 \]
Now, divide the entire equation by 2:
\[ b^2 + c^2 + 3b + 3c = 34 \]

### Step 5: Rearranging
Rearranging gives us:
\[ b^2 + c^2 + 3b + 3c - 34 = 0 \]

### Step 6: Finding \( ab - bc + ca \)
We can express \( ab - bc + ca \) using \( a = b + c + 3 \):
\[ ab - bc + ca = b(b + c + 3) - bc + c(b + c + 3) \]
Expanding this:
\[ = b^2 + bc + 3b - bc + cb + c^2 + 3c \]
This simplifies to:
\[ = b^2 + c^2 + 3b + 3c \]

### Step 7: Substitute back
From our earlier equation, we know:
\[ b^2 + c^2 + 3b + 3c = 34 \]
Thus,
\[ ab - bc + ca = 34 \]

### Final Answer
The value of \( ab - bc + ca \) is \( \boxed{34} \).

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