For all nonzero numbers x and y, the operation `grad` is defined by the equation `x grad y = (|x|)/(x^(2))+(y^(2))/(|y|)` when `x gt y` and by the equation `x grad y = (|x|)/(y^(2))-(x^(2))/(|y|)` when `x le y`. If `x grad y lt 0`, then which of the following could be true ?
I. `x^(3)=y^(3)`
II. `(y+x)(y-x)gt 0`
III. `x-y gt 0`

Options

A. I only
B. II only
C. III only
D. I and II

Explanation

To solve the problem, we need to analyze the operation defined by `x grad y` and determine under what conditions it can be less than 0.

### Step-by-step Solution:

1. **Understanding the Operation**:
The operation `x grad y` is defined differently based on the relationship between `x` and `y`:
- If `x > y`:
\[ x \text{ grad } y = \frac{|x|}{x^2} + \frac{y^2}{|y|} \]
- If `x \leq y`:
\[ x \text{ grad } y = \frac{|x|}{y^2} - \frac{x^2}{|y|} \]

2. **Condition Given**:
We are given that `x grad y < 0`.

3. **Analyzing the Case `x > y`**:
If `x > y`, then:
\[ x \text{ grad } y = \frac{|x|}{x^2} + \frac{y^2}{|y|} \]
Both terms are positive (since `|x|` and `|y|` are positive for non-zero `x` and `y`), thus `x grad y` cannot be less than 0. Therefore, this case does not satisfy our condition.

4. **Analyzing the Case `x \leq y`**:
We now consider the case when `x \leq y`:
\[ x \text{ grad } y = \frac{|x|}{y^2} - \frac{x^2}{|y|} \]
For this expression to be less than 0, we need:
\[ \frac{|x|}{y^2} < \frac{x^2}{|y|} \]
Rearranging gives:
\[ |x| \cdot |y| < x^2 \cdot y^2 \]
This implies:
\[ |x| < x^2 \cdot \frac{|y|}{y^2} \]

5. **Evaluating the Statements**:
- **Statement I: \(x^3 = y^3\)**:
This implies \(x = y\) (since both are non-zero). If \(x = y\), then \(x \leq y\) holds true. Thus, this statement could be true.

- **Statement II: \((y+x)(y-x) > 0\)**:
This can be rewritten as \(y^2 - x^2 > 0\), which means \(y^2 > x^2\) or \(|y| > |x|\). This could hold true if \(x \leq y\). Thus, this statement could also be true.

- **Statement III: \(x - y > 0\)**:
This implies \(x > y\), which contradicts our earlier conclusion that \(x \leq y\). Therefore, this statement cannot be true.

6. **Conclusion**:
The statements that could be true are I and II.

### Final Answer:
The answer is that statements I and II could be true.