(i) ` 2x^(3) = sqrt(256)`
II.` y^(2) -9y+14 = 0`

Options

A. x = y or no relation can be established between x & y.
B. ` x lt y`
C. ` x le y`
D. ` x ge y`

Explanation

Let's solve the given equations step by step.

### Step 1: Solve the first equation \(2x^3 = \sqrt{256}\)

1. **Calculate \(\sqrt{256}\)**:
\[ \sqrt{256} = 16 \]

2. **Substitute back into the equation**:
\[ 2x^3 = 16 \]

3. **Divide both sides by 2**:
\[ x^3 = \frac{16}{2} = 8 \]

4. **Take the cube root of both sides**:
\[ x = \sqrt[3]{8} \]
Since \(8 = 2^3\), we have:
\[ x = 2 \]

### Step 2: Solve the second equation \(y^2 - 9y + 14 = 0\)

1. **Factor the quadratic equation**:
We need to find two numbers that multiply to \(14\) and add up to \(-9\). The numbers are \(-7\) and \(-2\).
\[ y^2 - 7y - 2y + 14 = 0 \]

2. **Group the terms**:
\[ (y^2 - 7y) + (-2y + 14) = 0 \]

3. **Factor by grouping**:
\[ y(y - 7) - 2(y - 7) = 0 \]
\[ (y - 7)(y - 2) = 0 \]

4. **Set each factor to zero**:
\[ y - 7 = 0 \quad \text{or} \quad y - 2 = 0 \]
Thus, we find:
\[ y = 7 \quad \text{or} \quad y = 2 \]

### Step 3: Establish a relationship between \(x\) and \(y\)

1. **Values obtained**:
- From the first equation, \(x = 2\).
- From the second equation, \(y = 2\) or \(y = 7\).

2. **Compare values**:
- If \(y = 2\), then \(x = y\).
- If \(y = 7\), then \(x < y\) since \(2 < 7\).

### Conclusion

The relationship can be summarized as:
\[ x \leq y \]
Thus, the final answer is:
\[ x \text{ is less than or equal to } y \]

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