To find the value of \(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}\) given that \(x + y + z = 0\), we can use algebraic manipulation and properties of symmetric polynomial expressions. Here's the detailed, step-by-step solution:
[tex]\[
\text{Let} \; S = \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}.
\][/tex]
First, let's manipulate the expression:
[tex]\[
S = \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}.
\][/tex]
Rewrite each fraction with a common denominator:
[tex]\[
S = \frac{x^3}{xyz} + \frac{y^3}{xyz} + \frac{z^3}{xyz}.
\][/tex]
Combine the terms over a common denominator:
[tex]\[
S = \frac{x^3 + y^3 + z^3}{xyz}.
\][/tex]
Given that \(x + y + z = 0\), we can use a known algebraic identity for the sum of cubes:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx).
\][/tex]
Since \(x + y + z = 0\), the identity simplifies:
[tex]\[
x^3 + y^3 + z^3 - 3xyz = 0.
\][/tex]
Therefore, we have:
[tex]\[
x^3 + y^3 + z^3 = 3xyz.
\][/tex]
Substituting this result back into our expression for \(S\):
[tex]\[
S = \frac{x^3 + y^3 + z^3}{xyz} = \frac{3xyz}{xyz}.
\][/tex]
Simplify the fraction:
[tex]\[
S = 3.
\][/tex]
Thus, the value of \(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}\) is:
[tex]\[
\boxed{3}.
\][/tex]
