If the complex numbers `x and y` satisfy `x^3-y^3=98i and x-y=7i ,then x y=a+i b ,where a ,b , in Rdot` The value of `(a+b)//3` equals ______.

Explanation

To solve the problem, we start with the equations given:

1. \( x^3 - y^3 = 98i \)
2. \( x - y = 7i \)

We want to find \( xy = a + ib \) where \( a, b \in \mathbb{R} \) and then compute \( \frac{a + b}{3} \).

### Step 1: Use the identity for the difference of cubes

Recall the identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Substituting the known values:
\[ 98i = (7i)(x^2 + xy + y^2) \]

### Step 2: Simplify the equation

Dividing both sides by \( 7i \):
\[ x^2 + xy + y^2 = \frac{98i}{7i} = 14 \]

### Step 3: Express \( x^2 + y^2 \) in terms of \( xy \)

We know that:
\[ x^2 + y^2 = (x - y)^2 + 2xy \]
Substituting \( x - y = 7i \):
\[ x^2 + y^2 = (7i)^2 + 2xy = -49 + 2xy \]

### Step 4: Substitute into the equation

Now substitute \( x^2 + y^2 \) into the earlier equation:
\[ -49 + 2xy + xy = 14 \]
This simplifies to:
\[ -49 + 3xy = 14 \]

### Step 5: Solve for \( xy \)

Rearranging gives:
\[ 3xy = 14 + 49 \]
\[ 3xy = 63 \]
\[ xy = 21 \]

### Step 6: Identify \( a \) and \( b \)

Since \( xy = a + ib \), we have:
\[ a + ib = 21 + 0i \]
Thus, \( a = 21 \) and \( b = 0 \).

### Step 7: Calculate \( \frac{a + b}{3} \)

Now we compute:
\[ \frac{a + b}{3} = \frac{21 + 0}{3} = \frac{21}{3} = 7 \]

### Final Answer

The value of \( \frac{a + b}{3} \) equals **7**.

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