If `(x+y)^2-xy=0`, then find the value of `(x^3-y^3)//(x-y)` .
A. 2
B. 3
C. 0
D. 5

Options

A. C
B. A
C. B
D. D

Answer

To solve the equation \((x+y)^2 - xy = 0\) and find the value of \(\frac{x^3 - y^3}{x - y}\), we can follow these steps:

### Step 1: Simplify the given equation
We start with the equation:
\[ (x+y)^2 - xy = 0 \]
This can be rearranged to:
\[ (x+y)^2 = xy \]

### Step 2: Expand the left side
Expanding \((x+y)^2\) gives:
\[ x^2 + 2xy + y^2 = xy \]

### Step 3: Rearrange the equation
Now, we can rearrange this equation:
\[ x^2 + 2xy + y^2 - xy = 0 \]
This simplifies to:
\[ x^2 + xy + y^2 = 0 \]

### Step 4: Factor the expression
Notice that the expression \(x^2 + xy + y^2\) can be factored or analyzed. However, we can also use the identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Thus, we can write:
\[ \frac{x^3 - y^3}{x - y} = x^2 + xy + y^2 \]

### Step 5: Substitute the value from the earlier step
From our earlier step, we found that:
\[ x^2 + xy + y^2 = 0 \]
So, substituting this into our expression gives:
\[ \frac{x^3 - y^3}{x - y} = 0 \]

### Conclusion
Thus, the value of \(\frac{x^3 - y^3}{x - y}\) is:
\[ \boxed{0} \]