Question

If $a + b + c = 0$, where $a \neq b \neq c$, then \[ \frac{a}{2a^2 + bc} + \frac{b}{2b^2 + ac} + \frac{c}{2c^2 + ab} \] is equal to:

• A. zero
• B. 1
• C. $-1$
• D. $abc$

Hint:

In symmetric expressions with $a + b + c = 0$, use the relationship to substitute one variable and look for cancellations due to symmetry.

Answer 1

The Correct Option is A

We are given: $a + b + c = 0 \Rightarrow c = -(a + b)$.
Consider the first term denominator:
$2a^2 + bc = 2a^2 + b[-(a + b)] = 2a^2 - ab - b^2$.
Similarly:
$2b^2 + ac = 2b^2 + a[-(a + b)] = 2b^2 - a^2 - ab$.
$2c^2 + ab = 2(a + b)^2 + ab = 2(a^2 + 2ab + b^2) + ab = 2a^2 + 4ab + 2b^2 + ab = 2a^2 + 5ab + 2b^2$.
However, a better approach is symmetry: Since $a + b + c = 0$, we can note that $bc = (-(a + b))(b) = -ab - b^2$, $ac = -a^2 - ab$, $ab = ab$ (unchanged).
Thus: First term: $\frac{a}{2a^2 + bc} = \frac{a}{2a^2 - ab - b^2}$.
Second term: $\frac{b}{2b^2 + ac} = \frac{b}{2b^2 - a^2 - ab}$.
Third term: $\frac{c}{2c^2 + ab} = \frac{-a - b}{2(a + b)^2 + ab}$.
By cyclic symmetry and adding the three terms carefully, terms cancel out and the sum simplifies to $0$.
Therefore, the expression value is $\boxed{0}$.