If `a + b- c = 4 and a^(2) + b^(2) + c^(2)= 38`, find `ab- bc - ca`

Answer

To solve the problem, we start with the two equations given:

1. \( a + b - c = 4 \)
2. \( a^2 + b^2 + c^2 = 38 \)

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

### Step 1: Square the first equation

We start by squaring the first equation:

\[ (a + b - c)^2 = 4^2 \]

This simplifies to:

\[ a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = 16 \]

### Step 2: Substitute the second equation

Now, we can substitute the value of \( a^2 + b^2 + c^2 \) from the second equation into our squared equation:

\[ 38 + 2ab - 2bc - 2ca = 16 \]

### Step 3: Rearrange the equation

Next, we rearrange the equation to isolate the terms involving \( ab \), \( bc \), and \( ca \):

\[ 2ab - 2bc - 2ca = 16 - 38 \]

This simplifies to:

\[ 2ab - 2bc - 2ca = -22 \]

### Step 4: Divide by 2

To find \( ab - bc - ca \), we divide the entire equation by 2:

\[ ab - bc - ca = \frac{-22}{2} \]

This gives us:

\[ ab - bc - ca = -11 \]

### Final Answer

Thus, the value of \( ab - bc - ca \) is:

\[ \boxed{-11} \]

---