To solve the given problem, we need to find the value of the expression:
\(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}\)
given that \(x + y + z = 0\).
We can rewrite this condition as:
\(z = -(x + y)\)
Substituting \(z = -(x + y)\) into the expression gives us:
\(\frac{x^2}{y(-x-y)} + \frac{y^2}{x(-x-y)} + \frac{(-x-y)^2}{xy}\)
Simplifying each fraction:
\(\frac{x^2}{y(-x-y)} = -\frac{x^2}{y(x+y)}\)
\(\frac{y^2}{x(-x-y)} = -\frac{y^2}{x(x+y)}\)
\(\frac{(-x-y)^2}{xy} = \frac{x^2 + 2xy + y^2}{xy}\)
Thus, the expression can be rewritten as:
\(-\frac{x^2}{y(x+y)} - \frac{y^2}{x(x+y)} + \frac{x^2 + 2xy + y^2}{xy}\)
We can combine the first two terms:
\(-\left(\frac{x^3}{xy(x+y)} + \frac{y^3}{xy(x+y)}\right) = -\frac{x^3 + y^3}{xy(x+y)}\)
Using the identity \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\), and since \(z = -(x+y)\), we get:
\(x^3 + y^3 = -(x^2 - xy + y^2)(x+y)\)
So, the expression becomes:
\(\frac{x^2 + 2xy + y^2 - (x^2 - xy + y^2)}{xy} = 3\)
Hence, the final value of the expression is 3.
The correct option is 3.
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