Question

factorise the following:
64(a)^6-(b)^6

Answer 1

Hey Juhi !

Here is your solution :

= 64(a)^6 - (b)^6

= ( 2 )^6 ( a )^6 - ( b )^6

Using identity :

[ a^m × b^m = ( ab )^m ]

= ( 2a )^6 - ( b )^6

= [ ( 2a )^3 ]^2 - [ b^3 ]^2

Using identity :

[ a^2 - b^2 = ( a + b ) ( a - b ) ]

= [ ( 2a )^3 + b^3 ] [ ( 2a )^3 - b^3 ]

Using identity :

[ a^3 + b^3 = ( a + b ) ( a^2 + b^2 - ab )

= ( 2a + b ) { ( 2a )^2 + b^2 - 2a × b } [ ( 2a )^3 - b^3 ]


= ( 2a + b ) ( 4a^2 + b^2 - 2ab ) [ ( 2a )^3 - b^3 ]

Using identity :

[ a^3 - b^3 = ( a - b ) ( a^2 + b^2 + ab )

= ( 2a + b ) ( 4a^2 + b^2 - 2ab ) ( 2a - b ) [ ( 2a )^2 + b^2 + 2a × b ]

= ( 2a + b ) ( 4a^2 + b^2 - 2ab ) ( 2a - b ) ( 4a^2 + b^2 + 2ab )

Arranging the terms ,

= ( 2a + b ) ( 2a - b ) ( 4a^2 + b^2 - 2ab ) ( 4a^2 + b^2 + 2ab )


Hope it helps !!

Answer 2

Answer:

hey mate!

the answer with step by step solution is in the above attachment