Question

plz answer this question

Answer 1

Here's your answer :-
2^7x - 128x

128x - 128x = 0
Hope this helps
@PoojaBBSR
Sorry, since my 1st answer was actually from some other source... though I knew the answer, I took a screenshot & posted the answer since I was getting late for my dance classes... I'm so sorry (ーー")

Answer 2

= 2x⁷ - 128 x

By taking 2x common ,

= 2x ( x⁶ - 64 )

= 2x { ( x³ )² - ( 8 )² }

= 2x ( x³ + 8 ) ( x³ - 8 )    [ a² - b² = ( a + b ) ( a - b ) ]

= 2x { ( x )³ + ( 2 )³ }   { ( x )³ - ( 2 )³ }

= 2x ( x + 2 ) ( x² + 2² - 2x ) ( x - 2 ) ( x² + 2² + 2x )

= 2x ( x + 2 ) ( x - 2 ) ( x² + 4 - 2x ) ( x² + 4 + 2x ).

 [ a³ + b³ = ( a + b ) ( a² + b² - ab ) and ( a³ - b³ ) = ( a - b ) ( a² + b² + ab ) ]

Another method :

= 2x⁷ - 128x

By taking 2x common ,

= 2x ( x⁶ - 64 )

= 2x { ( x² )³ - ( 4 )³ }

= 2x ( x² - 4 ) { ( x²)² + (4)² + 4x }   [ a³ - b³ = ( a - b ) ( a² + b² + ab ) ]

= 2x { (x)² - (2)² }  ( x⁴ + 16 + 4x )

= 2x ( x + 2 ) ( x - 2 ) ( x⁴ + 16 + 4x ).