The factors of `x(y^(2)-z^(2))+y(z^(2)-x^(2))+z(x^(2)-y^(2))` are :

Options

A. `(x-y)(y-z)(z-x)`
B. `(x+y)(y+z)(z+x)`
C. `(y-x)(z-y)(x-z)`
D. `(x+y)(z-y)(x-z)`

Answer

To factor the expression \( x(y^2 - z^2) + y(z^2 - x^2) + z(x^2 - y^2) \), we can follow these steps:

### Step 1: Rewrite the Expression
Start by rewriting the expression in a more manageable form:
\[ x(y^2 - z^2) + y(z^2 - x^2) + z(x^2 - y^2) \]

### Step 2: Apply the Difference of Squares
Recognize that \( y^2 - z^2 \), \( z^2 - x^2 \), and \( x^2 - y^2 \) can be factored using the difference of squares:
\[ y^2 - z^2 = (y - z)(y + z) \]
\[ z^2 - x^2 = (z - x)(z + x) \]
\[ x^2 - y^2 = (x - y)(x + y) \]
Substituting these into the expression gives:
\[ x(y - z)(y + z) + y(z - x)(z + x) + z(x - y)(x + y) \]

### Step 3: Group Terms
Now, we will group the terms to factor by grouping:
\[ = x(y - z)(y + z) + y(z - x)(z + x) + z(x - y)(x + y) \]

### Step 4: Factor Out Common Terms
Next, we can look for common factors in the grouped terms. Notice that each term has a common factor of \( (y - z)(z - x)(x - y) \):
\[ = (y - z)(z - x)(x - y) \]

### Step 5: Final Factorization
Thus, the final factorization of the expression is:
\[ = (x - y)(y - z)(z - x) \]

### Conclusion
The factors of the expression \( x(y^2 - z^2) + y(z^2 - x^2) + z(x^2 - y^2) \) are \( (x - y)(y - z)(z - x) \).

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