Answer:
Sure, let's prove each of these identities:
i) \( (a+b)^2 = a^2 + 2ab + b^2 \)
Expanding \( (a+b)^2 \):
\[ (a+b)(a+b) = a(a+b) + b(a+b) \]
\[ = a^2 + ab + ba + b^2 \]
\[ = a^2 + 2ab + b^2 \]
ii) \( (a-b)^2 = a^2 - 2ab + b^2 \)
Expanding \( (a-b)^2 \):
\[ (a-b)(a-b) = a(a-b) - b(a-b) \]
\[ = a^2 - ab - ba + b^2 \]
\[ = a^2 - 2ab + b^2 \]
iii) \( a^2 - b^2 = (a+b)(a-b) \)
Expanding \( (a+b)(a-b) \):
\[ (a+b)(a-b) = a(a-b) + b(a-b) \]
\[ = a^2 - ab + ba - b^2 \]
\[ = a^2 - b^2 \]
Therefore, all three identities are proved.
please mark me as brainalist
