Let N be the set of natural numbers and a relation R on N be defined by
`R = { (x, y) in N xx N : x^(3) - 3 x ^(2) y - xy ^(2) + 3y ^(3) = 0 }` . Then the relation R is :

Options

A. symmetric but neither reflexive nor transitive
B. reflexive but neither symmetric nor transitive
C. reflexive and symmetric, but not transitive
D. an equivalence relation

Answer

To determine the nature of the relation \( R \) defined by the equation

\[ x^3 - 3x^2y - xy^2 + 3y^3 = 0, \]

we will analyze whether this relation is reflexive, symmetric, or transitive.

### Step 1: Factor the equation

We start with the equation:

\[ x^3 - 3x^2y - xy^2 + 3y^3 = 0. \]

We can factor this expression. First, we can group the terms:

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

Now, we can factor out \( (x - 3y) \):

\[ (x^2 - y^2)(x - 3y) = 0. \]

Using the difference of squares, we can further factor \( x^2 - y^2 \):

\[ (x + y)(x - y)(x - 3y) = 0. \]

### Step 2: Identify the conditions

From the factored equation, we have three cases:

1. \( x + y = 0 \)
2. \( x - y = 0 \) (which implies \( x = y \))
3. \( x - 3y = 0 \) (which implies \( x = 3y \))

### Step 3: Check reflexivity

A relation is reflexive if for every \( x \in N \), \( (x, x) \in R \).

For \( x = y \):

\[ x^3 - 3x^2x - xx^2 + 3x^3 = 0 \implies 0 = 0, \]

which is true. Therefore, \( R \) is reflexive.

### Step 4: Check symmetry

A relation is symmetric if whenever \( (x, y) \in R \), then \( (y, x) \in R \).

Consider \( x = 3y \):

Substituting \( x = 3y \) into the relation:

\[ (3y)^3 - 3(3y)^2y - (3y)y^2 + 3y^3 = 0. \]

Calculating this gives:

\[ 27y^3 - 27y^3 - 3y^3 + 3y^3 = 0 \implies 0 = 0, \]

which holds. However, consider \( (y, 3y) \):

Substituting \( y \) and \( 3y \):

\[ y^3 - 3y^2(3y) - y(3y)^2 + 3(3y)^3 = 0, \]

which simplifies to:

\[ y^3 - 9y^3 - 9y^3 + 81y^3 = 0 \implies 63y^3 = 0, \]

which does not hold for \( y \in N \). Thus, \( R \) is not symmetric.

### Step 5: Check transitivity

A relation is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \).

Consider \( x = 3y \) and \( y = 3z \):

Then \( x = 3(3z) = 9z \).

Check if \( (9z, z) \in R \):

\[ (9z)^3 - 3(9z)^2z - (9z)z^2 + 3z^3 = 0. \]

Calculating this gives:

\[ 729z^3 - 243z^3 - 9z^3 + 3z^3 = 0 \implies 480z^3 = 0, \]

which does not hold for \( z \in N \). Thus, \( R \) is not transitive.

### Conclusion

The relation \( R \) is reflexive but neither symmetric nor transitive.

### Final Answer

The relation \( R \) is reflexive but neither symmetric nor transitive.

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