What is ` (( x ^(2) - y^(2))^(3) + ( y ^(2) - z^(2))^(3) + ( z^(2) - x^(2))^(3))/(( x - y )^(3) + ( y - z)^(3) + ( z - x)^(3))`

Options

A. `((x + y) ( x + z))/( ( x + z))`
B. ` ( x + y)^(3) ( y + z) ^(3) ( z + x)^(3)`
C. `( x + y) ( y + z) ( z + x)`
D. ` ( x + y) ( y + z) `

Answer

To solve the expression

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

we can follow these steps:

### Step 1: Identify the components
We have two parts in the expression: the numerator and the denominator.

- The numerator consists of three cubes: \((x^2 - y^2)^3\), \((y^2 - z^2)^3\), and \((z^2 - x^2)^3\).
- The denominator consists of three cubes: \((x - y)^3\), \((y - z)^3\), and \((z - x)^3\).

### Step 2: Use the identity for sums of cubes
Recall the identity for the sum of cubes:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

This identity can be simplified when \(a + b + c = 0\), leading to:

\[ a^3 + b^3 + c^3 = 3abc \]

### Step 3: Check if the conditions are satisfied
For the numerator, we can set:

- \(a = x^2 - y^2\)
- \(b = y^2 - z^2\)
- \(c = z^2 - x^2\)

Now, we check if \(a + b + c = 0\):

\[ (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0 \]

This is indeed true, so we can apply the identity:

\[ (x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2) \]

### Step 4: Apply the identity to the denominator
Similarly, for the denominator, we can set:

- \(a = x - y\)
- \(b = y - z\)
- \(c = z - x\)

Again, we check if \(a + b + c = 0\):

\[ (x - y) + (y - z) + (z - x) = 0 \]

This is also true, so we can apply the identity:

\[ (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) \]

### Step 5: Substitute back into the expression
Now substituting back into our expression for \(L\):

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

### Step 6: Simplify
The \(3\) cancels out:

\[ L = \frac{(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{(x - y)(y - z)(z - x)} \]

### Step 7: Factor the differences of squares
Using the difference of squares, we can rewrite \(x^2 - y^2\) as \((x - y)(x + y)\), and similarly for the other terms:

\[ L = \frac{(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{(x - y)(y - z)(z - x)} \]

### Step 8: Cancel out common terms
The common terms \((x - y)\), \((y - z)\), and \((z - x)\) cancel out:

\[ L = (x + y)(y + z)(z + x) \]

### Final Result
Thus, the final simplified expression for \(L\) is:

\[ L = (x + y)(y + z)(z + x) \]

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