Question

Write the geometrical proof of (a+b)3.

Answer 1

HEY MATE HERE IS YOUR ANSWER

This can also be written as:
= (a + b) (a + b) (a + b) 


Now, multiply first two binomials as shown below: 
= { a(a + b) + b(a + b) } (a + b) 

= { a2 + ab + ab + b2 } (a + b) 

On rearranging the terms in curly braces we get: 
= { a2 + b2 + ab + ab } (a + b) 

= { a2 + b2 + 2ab } (a + b) 

Now, multiply trinomial with binomial as shown below: 
= a2(a + b) + b2(a + b) + 2ab(a + b) 

= a3 + a2b + ab2 + b3 + 2ab(a + b) 

= a3 + b3 + a2b + ab2 + 2ab(a + b) 
= a3 + b3 + a2b + ab2 + 2ab(a + b) 

Take ab common from the above red highlighted terms and we get: 
= a3 + b3 + ab(a + b) + 2ab(a + b) 

On adding like terms and we get: 
= a3 + b3 + 3ab(a + b) 

On solving it further we get: 
a3 + b3 + 3a2b + 3ab2 

Hence, in this way we obtain the identity i.e. (a + b)3 = a3 + b3 + 3ab(a + b) = a3 + b3+ 3a2b + 3ab2 

HOPE THIS HELPS YOU......✌✌

Answer 2

In the below figure, the cube has each side as (a+b) where the length of blue line on each side is a and the length of red line on each side is b.

Volume of the cube =$(a+b)^3$

This cube is actually formed by the following 8 pieces

Piece 1 -> cube with each side 'a'.

Volume of piece 1= a×a×a=$a^3$  

Piece 2 -> cube with each side 'b'

Volume of piece 2 =b×b×b=$b^3$

Piece 3, Piece 5 and Piece 7 are cuboids of sides a, b, b

So, Volume of Piece 3 = a×b×b=$ab^2$

Sum of Volume s of Piece 3, Piece 5 and Piece 7 = $ab^2+ab^2+ab^2=3ab^2$

Piece 4, Piece 6 and Piece 8 are cuboids of sides a, a, b

So, Volume of Piece 4 = a×a×b=$a^2b$

Sum of Volume s of Piece 4, Piece 6 and Piece 8 = $a^2b+a^2b+a^2b=3a^2b$

Summing all these volumes will give the volume of the cube of side (a+b)

$(a+b)^3=a^3+b^3+3ab^2+3a^2b$

Thus proved.