`(x^3 + y^3 + z^3 - 3xyz)/(a^3 + b^3 + c^3 - 3abc) xx (a^2 + b^2 + c^2 - ab - bc - ca)/(x^2 + y^2 + z^2 - xy - yz - zx)` equals

Options

A. `1`
B. `(x^2 + y^2 +z^2)/(a^2 + b^2 + c^2)`
C. `(x+y+z)/(a+b+c )`
D. `(xyz)/(abc)`

Answer

To simplify the expression \((x^3 + y^3 + z^3 - 3xyz)/(a^3 + b^3 + c^3 - 3abc) \times (a^2 + b^2 + c^2 - ab - bc - ca)/(x^2 + y^2 + z^2 - xy - yz - zx)\), we can use the identity for the sum of cubes.

### Step-by-Step Solution:

1. **Identify the Identity**:
We know that:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
This identity will help us simplify both the numerator and denominator of the first fraction.

2. **Apply the Identity to the Numerator**:
For the numerator \(x^3 + y^3 + z^3 - 3xyz\), we can apply the same identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]

3. **Substitute into the Expression**:
Substitute the identities into the original expression:
\[ \frac{(x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz)}{(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)} \times \frac{(a^2 + b^2 + c^2 - ab - ac - bc)}{(x^2 + y^2 + z^2 - xy - xz - yz)} \]

4. **Cancel Out Common Terms**:
In the expression, we can see that \( (x^2 + y^2 + z^2 - xy - xz - yz) \) in the numerator of the first fraction cancels with the same term in the denominator of the second fraction. Similarly, \( (a^2 + b^2 + c^2 - ab - ac - bc) \) cancels out:
\[ \frac{(x + y + z)}{(a + b + c)} \]

5. **Final Result**:
Thus, the simplified expression is:
\[ \frac{x + y + z}{a + b + c} \]