Positive real numbers $a$ and $b$ verify $a^5+b^5=a^3+b^3$. Find the greatest possible value of the expression $E=a^2-ab+b^2$.
1. Given the equation \(a^5 + b^5 = a^3 + b^3\), we start by setting \(k = \frac{a}{b}\). This implies \(a = kb\).

2. Substitute \(a = kb\) into the given equation:
   \[
   (kb)^5 + b^5 = (kb)^3 + b^3
   \]
   Simplifying, we get:
   \[
   k^5 b^5 + b^5 = k^3 b^3 + b^3
   \]
   Factor out \(b^5\) and \(b^3\) respectively:
   \[
   b^5 (k^5 + 1) = b^3 (k^3 + 1)
   \]
   Since \(b \neq 0\), we can divide both sides by \(b^3\):
   \[
   b^2 (k^5 + 1) = k^3 + 1
   \]
   Solving for \(b^2\), we get:
   \[
   b^2 = \frac{k^3 + 1}{k^5 + 1}
   \]

3. Now, we need to find the greatest possible value of the expression \(E = a^2 - ab + b^2\). Substitute \(a = kb\) into \(E\):
   \[
   E = (kb)^2 - (kb)b + b^2 = k^2 b^2 - kb^2 + b^2
   \]
   Factor out \(b^2\):
   \[
   E = b^2 (k^2 - k + 1)
   \]

4. Substitute \(b^2 = \frac{k^3 + 1}{k^5 + 1}\) into \(E\):
   \[
   E = \frac{k^3 + 1}{k^5 + 1} (k^2 - k + 1)
   \]

5. To find the maximum value of \(E\), we analyze the function:
   \[
   E = \frac{(k^3 + 1)(k^2 - k + 1)}{k^5 + 1}
   \]

6. We use the Cauchy-Schwarz inequality to show that \(E \leq 1\). The Cauchy-Schwarz inequality states that for any real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\):
   \[
   (x_1^2 + x_2^2 + \cdots + x_n^2)(y_1^2 + y_2^2 + \cdots + y_n^2) \geq (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2
   \]

7. Applying this to our problem, we set \(x_1 = a^2\), \(x_2 = b^2\), \(y_1 = 1\), and \(y_2 = 1\):
   \[
   (a^2 + b^2)(1 + 1) \geq (a + b)^2
   \]
   Simplifying, we get:
   \[
   2(a^2 + b^2) \geq (a + b)^2
   \]
   \[
   2(a^2 + b^2) \geq a^2 + 2ab + b^2
   \]
   \[
   a^2 + b^2 \geq 2ab
   \]
   \[
   a^2 - ab + b^2 \geq ab
   \]

8. Since \(a = b = 1\) satisfies the given equation \(a^5 + b^5 = a^3 + b^3\), we substitute \(a = 1\) and \(b = 1\) into \(E\):
   \[
   E = 1^2 - 1 \cdot 1 + 1^2 = 1 - 1 + 1 = 1
   \]

9. Therefore, the greatest possible value of \(E\) is indeed \(1\).

The final answer is \(\boxed{1}\).