If `x^(2)+y^(2)=4`, then maximum value of `((x^(3)+y^(3))/(x+y))` is

Explanation

To find the maximum value of \(\frac{x^3 + y^3}{x + y}\) given that \(x^2 + y^2 = 4\), we can follow these steps:

### Step 1: Rewrite \(x^3 + y^3\)
Using the identity for the sum of cubes, we have:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
Thus,
\[ \frac{x^3 + y^3}{x + y} = x^2 - xy + y^2 \]

### Step 2: Substitute \(x^2 + y^2\)
From the given condition \(x^2 + y^2 = 4\), we can substitute this into our expression:
\[ x^2 - xy + y^2 = (x^2 + y^2) - xy = 4 - xy \]

### Step 3: Express \(xy\) in terms of \(x^2 + y^2\)
We know from the identity:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
Let \(s = x + y\) and \(p = xy\). Then we can express \(p\) as:
\[ p = \frac{s^2 - 4}{2} \]
Substituting this into our expression gives:
\[ 4 - xy = 4 - p = 4 - \frac{s^2 - 4}{2} = 4 - \frac{s^2}{2} + 2 = 6 - \frac{s^2}{2} \]

### Step 4: Maximize the expression
To maximize \(6 - \frac{s^2}{2}\), we need to find the maximum value of \(s^2\). From the condition \(x^2 + y^2 = 4\), we also know:
\[ s^2 = (x + y)^2 = x^2 + y^2 + 2xy \leq 4 + 2xy \]
Since \(x^2 + y^2 = 4\), we can also express \(xy\) in terms of \(s\):
\[ xy = \frac{s^2 - 4}{2} \]
To find the maximum value of \(s\), we consider the maximum value of \(xy\) which occurs when \(x\) and \(y\) are equal. Thus, the maximum occurs when \(x = y\).

### Step 5: Solve for \(x\) and \(y\)
Let \(x = y\). Then:
\[ 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2}, y = \sqrt{2} \]
Thus, \(s = x + y = 2\sqrt{2}\).

### Step 6: Substitute back to find the maximum value
Now substituting \(s = 2\sqrt{2}\) back into our expression:
\[ 6 - \frac{s^2}{2} = 6 - \frac{(2\sqrt{2})^2}{2} = 6 - \frac{8}{2} = 6 - 4 = 2 \]

### Conclusion
The maximum value of \(\frac{x^3 + y^3}{x + y}\) is:
\[ \boxed{6} \]