Question

Prove that (a + b)^2 = a^2 + b^2 + 2ab

Answer 1

Hii There!!


Here, consider L.H.S :

( a + b )^2 = ( a + b ) ( a + b )


=> a^2 + ab + ba + b^2


Since, ab = ba [ by commutative property ]


=> a^2 + 2ab + b^2 = R.H.S


So, ( a + b )^2 = a^2 + 2ab + b^2

Hence, verified.




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Hope it helps☺

Answer 2

Heya! Here's your answer....

$\text{ (a + b)^2 can be written as (a + b)(a + b) } \\ \\\text{so, (a + b)( a + b) = a^2 + ab + ba + b^2 \\ \\--(by Multiplication....)} \\ \\= a^2 + ab + ab + b^2 -(because \: a \times b = b \times a....) \\ \\= a^2 + 2ab + b^2 \\ \\ \boxed{\text{Therefore, (a + b)^2 = a^2 + 2ab + b^2 .}} \\ \\ Hence \: \:Proved.....$

Hope it helps!