Study the given statements carefully.
Statement I:
`((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a+b)^(3)+(b+c)^(3)+(c+a)^(3))`
`=(a+b)(b+c)(c+a)`
Statement II `a^(2)+b^(2)+c^(2)-ab-bc-ca`
`=1/2[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]`
Which of the following options holds?

Options

A. Both Statement I and Statement II are true.
B. Statement I is true but Statement II is false
C. Statement I is false but Statement II is true.
D. Both Statement I and Statement II are false.

Explanation

To solve the problem, we will analyze both statements step by step.

### Statement I:
We need to evaluate the expression:
\[ \frac{(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3}{(a+b)^3 + (b+c)^3 + (c+a)^3} \]
and check if it equals \((a+b)(b+c)(c+a)\).

1. **Recognize the identity for cubes**:
We know that \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)\). This holds when \(x + y + z = 0\).

2. **Set \(x = a^2 - b^2\), \(y = b^2 - c^2\), \(z = c^2 - a^2\)**:
Now, we check if \(x + y + z = 0\):
\[ (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0 \]
This condition holds.

3. **Apply the identity**:
Since \(x + y + z = 0\), we can use the identity:
\[ (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2) \]

4. **Evaluate the denominator**:
For the denominator, we have:
\[ (a+b)^3 + (b+c)^3 + (c+a)^3 \]
We can also apply the same identity here. Set \(p = a+b\), \(q = b+c\), \(r = c+a\). Then:
\[ p + q + r = (a+b) + (b+c) + (c+a) = 2(a+b+c) \]
This does not equal zero unless \(a + b + c = 0\), but we can still evaluate it directly.

5. **Final comparison**:
After evaluating both sides, we find that the left-hand side does not simplify to \((a+b)(b+c)(c+a)\) under general conditions. Thus, **Statement I is false**.

### Statement II:
We need to verify:
\[ a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \]

1. **Expand the right-hand side**:
\[ (a-b)^2 = a^2 - 2ab + b^2 \]
\[ (b-c)^2 = b^2 - 2bc + c^2 \]
\[ (c-a)^2 = c^2 - 2ca + a^2 \]
Adding these gives:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2a^2 + 2b^2 + 2c^2 - 2(ab + bc + ca) \]

2. **Multiply by \(\frac{1}{2}\)**:
\[ \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] = a^2 + b^2 + c^2 - (ab + bc + ca) \]
This matches the left-hand side.

3. **Conclusion**:
Since both sides are equal, **Statement II is true**.

### Final Conclusion:
- **Statement I is false**.
- **Statement II is true**.

Thus, the correct option is that Statement I is false and Statement II is true.