Question

27 and 28 question numbers do it if u can.......

Answer 1

Question 27 :

Given :

( x - 3 y )³ + ( 3 y - 7 z )³ + ( 7 z - x )³

These are in the form : a³ + b³ + c³

a³ + b³ + c³ = 3 abc when a + b + c = 0

So  :

( x - 3 y )³ + ( 3 y - 7 z )³ + ( 7 z - x )³

= > 3 ( x - 3 y )( 3 y - 7 z )( 7 z - x )

Question 28 :

x + y + z = 0

⇒ x + y =  - z ......( 1 )

⇒ ( x + y )³ = - z³

⇒ x³ + y³ + 3 xy ( x + y ) = - z³ [ from 1 put the value => ]

⇒ x³ + y³ - 3 xyz = - z³

⇒ x³ + y³ + z³ = 3 xyz [ PROVED ]