The equation `x^(3)+x^(2)y-xy^(2)-y^(3)=0` represents three straight lines passing through the origin such that

Options

A. two of them are coincident and two of them are perpendicular
B. two of them are coincident but not two are perpendicular
C. two of them are perpendicular but no two are coincident
D. none of these

Explanation

To solve the equation \(x^3 + x^2y - xy^2 - y^3 = 0\) and determine the nature of the straight lines it represents, we can follow these steps:

### Step 1: Factor the given equation
We start with the equation:
\[ x^3 + x^2y - xy^2 - y^3 = 0 \]
We can group the terms to factor them. Taking \(x^2\) common from the first two terms and \(-y^2\) from the last two terms gives:
\[ x^2(x + y) - y^2(x + y) = 0 \]
Now, we can factor out \((x + y)\):
\[ (x + y)(x^2 - y^2) = 0 \]

### Step 2: Further factor \(x^2 - y^2\)
The expression \(x^2 - y^2\) can be factored using the difference of squares:
\[ x^2 - y^2 = (x - y)(x + y) \]
Thus, we can rewrite the equation as:
\[ (x + y)(x - y)(x + y) = 0 \]
This simplifies to:
\[ (x + y)^2(x - y) = 0 \]

### Step 3: Identify the lines
From the factored form, we can identify the lines represented by the equation:
1. \(x + y = 0\) (This line is counted twice since it appears squared)
2. \(x - y = 0\)

Thus, we have two distinct lines: \(x + y = 0\) (which is coincident) and \(x - y = 0\).

### Step 4: Determine the slopes of the lines
Now, we need to find the slopes of these lines to check if they are perpendicular:
- For the line \(x + y = 0\):
\[ y = -x \quad \Rightarrow \quad m_1 = -1 \]
- For the line \(x - y = 0\):
\[ y = x \quad \Rightarrow \quad m_2 = 1 \]

### Step 5: Check for perpendicularity
To check if the lines are perpendicular, we calculate the product of their slopes:
\[ m_1 \cdot m_2 = (-1) \cdot (1) = -1 \]
Since the product of the slopes is \(-1\), it confirms that the lines are perpendicular.

### Conclusion
The equation \(x^3 + x^2y - xy^2 - y^3 = 0\) represents three lines passing through the origin, where two lines are coincident and one line is perpendicular to them.