If `a+b+c=0`, then what is the value of `(a^(2)+b^(2)+c^(2))/((a-b)^(2)+(b-c)^(2)+(c-a)^(2))`?

Options

A. 1
B. 3
C. `(1)/(3)`
D. 0

Answer

To solve the problem, we need to find the value of

\[ \frac{a^2 + b^2 + c^2}{(a-b)^2 + (b-c)^2 + (c-a)^2} \]

given that \( a + b + c = 0 \).

### Step 1: Use the condition \( a + b + c = 0 \)

Since \( a + b + c = 0 \), we can express \( c \) in terms of \( a \) and \( b \):

\[ c = - (a + b) \]

### Step 2: Substitute \( c \) into \( a^2 + b^2 + c^2 \)

Now, we substitute \( c \) into the expression \( a^2 + b^2 + c^2 \):

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

Thus,

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

### Step 3: Expand the denominator \((a-b)^2 + (b-c)^2 + (c-a)^2\)

Now, we need to expand the denominator:

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

Now, adding these:

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

Combining like terms:

\[ = (a^2 + a^2 + 4a^2) + (b^2 + 4b^2 + b^2) + (-2ab + 4ab + 4ab) \]
\[ = 6a^2 + 6b^2 + 6ab = 6(a^2 + b^2 + ab) \]

### Step 4: Substitute back into the original expression

Now, we can substitute back into our original expression:

\[ \frac{a^2 + b^2 + c^2}{(a-b)^2 + (b-c)^2 + (c-a)^2} = \frac{2(a^2 + b^2 + ab)}{6(a^2 + b^2 + ab)} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer

Thus, the value of

\[ \frac{a^2 + b^2 + c^2}{(a-b)^2 + (b-c)^2 + (c-a)^2} = \frac{1}{3} \]