Explanation:

An AVL tree is a self-balancing binary search tree in which the heights of the left and right subtrees of any node differ by at most one. To maintain this balance, rotations are performed when necessary during insertion or deletion operations.

Step 1: Insertion of 9
- Since the AVL tree is initially empty, we simply insert the element 9 as the root of the tree. No rotations are required at this point.

Step 2: Insertion of 6
- The element 6 is less than 9, so it becomes the left child of 9.
- The balance factor of the node 9 becomes -1 (left subtree height - right subtree height).
- Since the balance factor of 9 is not within the range [-1, 1], a rotation is required to balance the tree.
- We perform a right rotation on the node 9, making 6 the new root and 9 its right child.
- The balance factors of 6 and 9 become 0 and 0, respectively.
- The resulting tree after the rotation is:
```
6
\
9
```

Step 3: Insertion of 5
- The element 5 is less than 6, so it becomes the left child of 6.
- The balance factor of the node 6 becomes 1 (left subtree height - right subtree height).
- Since the balance factor of 6 is not within the range [-1, 1], a rotation is required to balance the tree.
- We perform a right rotation on the node 6, making 5 the new root and 6 its right child.
- The balance factors of 5 and 6 become 0 and 0, respectively.
- The resulting tree after the rotation is:
```
5
\
6
\
9
```

Step 4: Insertion of 8
- The element 8 is greater than 5 and less than 6, so it becomes the right child of 5.
- The balance factor of the node 5 becomes -1 (left subtree height - right subtree height).
- Since the balance factor of 5 is not within the range [-1, 1], a rotation is required to balance the tree.
- We perform a left rotation on the node 5, making 6 the new root and 5 its left child.
- The balance factors of 5 and 6 become 0 and 0, respectively.
- The resulting tree after the rotation is:
```
6
/ \
5 9
\
8
```

Step 5: Insertion of 7
- The element 7 is greater than 6 and less than 9, so it becomes the right child of 6.
- The balance factor of the node 6 becomes -1 (left subtree height - right subtree height).
- Since the balance factor of 6 is not within the range [-1, 1], a rotation is required to balance the tree.
- We perform a left rotation on the node 6, making 7 the new root and 6 its left child.
- The balance factors of 6 and 7 become 0 and