Answer :

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]

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