If `a+b+c=11` and `ab+bc+ca=35`, find the value of `(a-b)^(2)+(b-c)^(2)+(c-a)^(2)`

Options

A. 24
B. 64
C. 16
D. 32

Answer

To solve the problem, we need to find the value of \((a-b)^2 + (b-c)^2 + (c-a)^2\) given that \(a + b + c = 11\) and \(ab + bc + ca = 35\).

### Step-by-Step Solution:

1. **Use the identity for the expression**:
We can use the identity:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2) - 2(ab + ac + bc) \]
This allows us to express the desired quantity in terms of \(a^2 + b^2 + c^2\) and \(ab + ac + bc\).

2. **Calculate \(a^2 + b^2 + c^2\)**:
We know:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting the known values:
\[ 11^2 = a^2 + b^2 + c^2 + 2(35) \]
This simplifies to:
\[ 121 = a^2 + b^2 + c^2 + 70 \]
Rearranging gives:
\[ a^2 + b^2 + c^2 = 121 - 70 = 51 \]

3. **Substitute values into the identity**:
Now we substitute \(a^2 + b^2 + c^2\) and \(ab + ac + bc\) into the identity:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(51) - 2(35) \]
This simplifies to:
\[ = 102 - 70 = 32 \]

4. **Final answer**:
Therefore, the value of \((a-b)^2 + (b-c)^2 + (c-a)^2\) is:
\[ \boxed{32} \]