TT 9.
(4 marks) 7. After completing this lesson, Bohdan made the following conclusions:
I. $a^{2}-2 a b-b^{2}=(a-b)^{2}$ as long as $b \neq 0$
II. $a^{2}+b^{2}=(a+b)(a+b)$
III. $a^{2}-b^{2}=a^{2}-2 a b+b^{2}$
IV. $(a+b)^{2}=a^{2}+2 a b+b^{2}$
Analyze his statements and state whether each equation is sometimes, always, or never true. Explain your reasoning for each.

Step 1 ：Analyze each statement:

Step 2 ：I. \(a^{2}-2 a b-b^{2}=(a-b)^{2}\) as long as \(b \neq 0\)

Step 3 ：Expand the right side: \((a-b)^{2} = a^2 - 2ab + b^2\)

Step 4 ：Compare with the left side: \(a^2 - 2ab - b^2\)

Step 5 ：This statement is \(\boxed{\text{never}}\) true.

Step 6 ：II. \(a^{2}+b^{2}=(a+b)(a+b)\)

Step 7 ：Expand the right side: \((a+b)(a+b) = a^2 + 2ab + b^2\)

Step 8 ：Compare with the left side: \(a^2 + b^2\)

Step 9 ：This statement is \(\boxed{\text{never}}\) true.

Step 10 ：III. \(a^{2}-b^{2}=a^{2}-2 a b+b^{2}\)

Step 11 ：Notice that \(a^2 - b^2 = (a+b)(a-b)\)

Step 12 ：Expand the right side: \(a^2 - 2ab + b^2\)

Step 13 ：This statement is \(\boxed{\text{never}}\) true.

Step 14 ：IV. \((a+b)^{2}=a^{2}+2 a b+b^{2}\)

Step 15 ：Expand the left side: \((a+b)^2 = a^2 + 2ab + b^2\)

Step 16 ：Compare with the right side: \(a^2 + 2ab + b^2\)

Step 17 ：This statement is \(\boxed{\text{always}}\) true.