Understanding the Algebra Problem with x+y+z=0

The problem asks for the value of a specific algebraic expression given a condition relating three variables, x, y, and z. The given condition is \(x + y + z = 0\).

The expression we need to evaluate is:

\[ \frac {x} {(yz)^2}+ \frac {y} {(xz)^2} + \frac {z} {(xy)^2} \]

Simplifying the Algebraic Expression

To combine these fractions, we need to find a common denominator. The denominators are \((yz)^2\), \((xz)^2\), and \((xy)^2\). The least common multiple of these denominators is \((xyz)^2\).

Let's rewrite each term with the common denominator \((xyz)^2\):

1. The first term is \(\frac {x} {(yz)^2}\). To get the denominator \((xyz)^2\), we multiply the numerator and denominator by \(x^2\): \[ \frac {x} {(yz)^2} = \frac {x \cdot x^2} {(yz)^2 \cdot x^2} = \frac {x^3} {(xyz)^2} \]
2. The second term is \(\frac {y} {(xz)^2}\). To get the denominator \((xyz)^2\), we multiply the numerator and denominator by \(y^2\): \[ \frac {y} {(xz)^2} = \frac {y \cdot y^2} {(xz)^2 \cdot y^2} = \frac {y^3} {(xyz)^2} \]
3. The third term is \(\frac {z} {(xy)^2}\). To get the denominator \((xyz)^2\), we multiply the numerator and denominator by \(z^2\): \[ \frac {z} {(xy)^2} = \frac {z \cdot z^2} {(xy)^2 \cdot z^2} = \frac {z^3} {(xyz)^2} \]

Now, we can add the terms with the common denominator:

\[ \frac {x^3} {(xyz)^2} + \frac {y^3} {(xyz)^2} + \frac {z^3} {(xyz)^2} = \frac {x^3 + y^3 + z^3} {(xyz)^2} \]

Using the Condition x+y+z=0

We are given the condition \(x + y + z = 0\). There is a useful algebraic identity related to this condition:

If \(x + y + z = 0\), then \(x^3 + y^3 + z^3 = 3xyz\).

We can substitute \(x^3 + y^3 + z^3\) with \(3xyz\) in our simplified expression:

\[ \frac {x^3 + y^3 + z^3} {(xyz)^2} = \frac {3xyz} {(xyz)^2} \]

Final Simplification

Now, we simplify the expression \(\frac {3xyz} {(xyz)^2}\):

\[ \frac {3xyz} {(xyz)^2} = \frac {3xyz} {x^2y^2z^2} \]

Assuming \(x, y, z \neq 0\) (otherwise the original expression would be undefined due to division by zero), we can cancel one power of \(xyz\) from the numerator and the denominator:

\[ \frac {3\cancel{xyz}} {x^2y^2z^2} = \frac {3} {xyz} \]

Thus, the value of the expression \(\frac {x} {(yz)^2}+ \frac {y} {(xz)^2} + \frac {z} {(xy)^2}\) when \(x + y + z = 0\) is \(\frac {3} {xyz}\).

Checking the Options

Let's compare our result with the given options:

1. \(\frac {xyz} {3}\)
2. \(\frac {3} {xyz}\)
3. xyz
4. \(\frac {1} {xyz}\)

Our calculated value matches option 2.

Revision Table: Key Concepts in Algebra

Understanding fundamental algebraic identities and how to manipulate expressions is crucial for solving such problems. This question relies on simplifying fractions and applying a specific identity that holds true under the condition \(x+y+z=0\).

Additional Information: Algebraic Identities

Algebraic identities are equations that are true for all values of the variables involved. The identity used here, \(x+y+z=0 \implies x^3+y^3+z^3=3xyz\), is derived from the more general identity \(x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). If \(x+y+z=0\), then the right side becomes \(0 \cdot (x^2+y^2+z^2-xy-yz-zx) = 0\), leading to \(x^3+y^3+z^3 - 3xyz = 0\), which means \(x^3+y^3+z^3 = 3xyz\). This identity is very useful in various algebraic problems.