Question

a + b ki power 2 equal to a2,+2ab+b2 prove that​

Answer 1

Answer:

WE SHOULD PROVE (a+b)^2=a^2+2ab+b^2

Step-by-step explanation:

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

(a+b)(a+b)

NOW MULTIPLYING,

a(a+b)+b(a+b)

a^2+ab+ba+b^2

THEREFORE,  a^2+2ab+b^2

     HENCE PROVED

Answer 2

AnswEr:-

To Prove:-

$\tt\longrightarrow {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} .$

So Here,

We will simply, simplify the LHS to get the result which is equal to RHS.

$\tt\therefore\:\: LHS=(a + b) {}^{2} . \\ \\ \tt = (a + b)( a + b). \\ \\ \tt= a(a+b)+b(a+b) \\ \\ \tt = ( a \times a) + (a \times b) + (a \times b) + (b \times b).\\ \\ \tt = ( {a}^{2}) + (ab) + ( ab) + ( {b})^{2} . \\ \\ \tt = {a}^{2} + ab + ab + {b}^{2} . \\ \\ \tt = {a}^{2} + 2ab + {b}^{2} . \\ \\ \tt = RHS$

$\therefore$ LHS = RHS.

Hence Proved.

$\rule{200}2$

More Algebraic IDs:-

$\\ \tt\longrightarrow (a-b)^2 = a^2-2ab+b^2. \\ \\ \tt\longrightarrow(a+b)(a-b) = a^2-b^2. \\ \\ \tt\longrightarrow (x+a)(x+b) = x^2 + (a+b)x +ab. \\ \\ \tt\longrightarrow a^2+b^2= (a+b)^2-2ab.$