If `x = 6, y = 4 and z = 3, ` then find the value of ` x ^(2) + y ^(2) + z ^(2) - 2xy + 2 yz - 2 zx.`

Options

A. 125
B. 1
C. `-25`
D. 10

Answer

To solve the expression \( x^2 + y^2 + z^2 - 2xy + 2yz - 2zx \) given \( x = 6, y = 4, z = 3 \), we will substitute the values and simplify step by step.

### Step-by-Step Solution:

1. **Substitute the values of x, y, and z into the expression:**
\[ x^2 + y^2 + z^2 - 2xy + 2yz - 2zx = 6^2 + 4^2 + 3^2 - 2(6)(4) + 2(4)(3) - 2(3)(6) \]

2. **Calculate the squares:**
\[ 6^2 = 36, \quad 4^2 = 16, \quad 3^2 = 9 \]
So, we have:
\[ 36 + 16 + 9 \]

3. **Sum the squares:**
\[ 36 + 16 + 9 = 61 \]

4. **Calculate the products:**
- For \( -2xy \):
\[ -2(6)(4) = -48 \]
- For \( +2yz \):
\[ 2(4)(3) = 24 \]
- For \( -2zx \):
\[ -2(3)(6) = -36 \]

5. **Combine all parts of the expression:**
\[ 61 - 48 + 24 - 36 \]

6. **Perform the arithmetic step by step:**
- First, calculate \( 61 - 48 \):
\[ 61 - 48 = 13 \]
- Next, add \( 24 \):
\[ 13 + 24 = 37 \]
- Finally, subtract \( 36 \):
\[ 37 - 36 = 1 \]

7. **Final result:**
\[ \text{The value of the expression is } 1. \]