Question

using suitable identity find (x+y+z)^2-(x-y-z)^2​

Answer 1

The answer is " 0 "

Step-by-step explanation:

This is in the form of a^2 - b^2

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

= [ ( x + y + z ) + ( - x - y - z ) ] [ ( x + y + z ) - ( - x - y - z ) ]

= ( 0 ) * [ (  x + y + z ) + ( x + y + z ) ]

= ( 0 ) * ( 2x + 2y + 2z )

= 0 .

( x + y + z )^2 - ( - x - y - z )^2 = 0  

Answer 2

Answer:

Step-by-step explanation:

( x + y + z )² - ( x - y - z )²

= x² + y² + z² + 2xy + 2yz + 2zx - ( x² + y² + z² - 2xy + 2yz - 2zx )

= x² + y² + z² + 2xy + 2yz + 2zx - x² - y² - z² + 2xy - 2yz + 2zx (∵ - × + = - ; - × - = + )

= 4xy + 4zx ( or ) 4x ( y + z ) is the answer.