If x + y + z = 22 and xy + yz + zx = 35 , then what is the value of `( x- y)^(2) + ( y - z)^(2) + ( z - x)^(2)` ?

Options

A. 793
B. 681
C. 758
D. 715

Answer

To solve the problem, we need to find the value of \( (x - y)^2 + (y - z)^2 + (z - x)^2 \) given the equations \( x + y + z = 22 \) and \( xy + yz + zx = 35 \).

### Step 1: Use the identity for the sum of squares
We can use the identity:
\[ (x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \]
This allows us to express the desired quantity in terms of \( x^2 + y^2 + z^2 \) and \( xy + yz + zx \).

### Step 2: Calculate \( x^2 + y^2 + z^2 \)
From the first equation, we know:
\[ x + y + z = 22 \]
Squaring both sides gives:
\[ (x + y + z)^2 = 22^2 = 484 \]
Expanding the left side, we have:
\[ x^2 + y^2 + z^2 + 2(xy + yz + zx) = 484 \]
Substituting \( xy + yz + zx = 35 \) into the equation:
\[ x^2 + y^2 + z^2 + 2 \cdot 35 = 484 \]
\[ x^2 + y^2 + z^2 + 70 = 484 \]
Now, solving for \( x^2 + y^2 + z^2 \):
\[ x^2 + y^2 + z^2 = 484 - 70 = 414 \]

### Step 3: Substitute into the identity
Now we can substitute \( x^2 + y^2 + z^2 \) and \( xy + yz + zx \) back into our identity:
\[ (x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \]
Substituting the values we found:
\[ = 2(414) - 2(35) \]
Calculating each part:
\[ = 828 - 70 = 758 \]

### Final Answer
Thus, the value of \( (x - y)^2 + (y - z)^2 + (z - x)^2 \) is \( \boxed{758} \).