Question

( 3k - 4r - 2m )² - (3k + 4r - 2m )²​

Answer 1

(3k - 4r - 2m)² - (3k + 4r - 2m)²

→ (3k)² - (4r - 2m)²

→ 9k² - (16r² + 4m² - 8rm)

→ 9k² - 16r² - 4m² + 8rm

Answer 2

Answer:

32mr - 48kr  

OR  16r (2m - 3k)   [ If we take 16r common]

Explanation:

To find: The value of ( 3k - 4r - 2m )² - (3k + 4r - 2m )²​

Proof:

= ( 3k - 4r - 2m )² - (3k + 4r - 2m )²​

= [ 3k - 4r - 2m + (3k + 4r - 2m) ] [ 3k - 4r - 2m - (3k + 4r - 2m)

Identity: $[ a^{2} - b^{2} = (a+b) (a-b) ]$

= (3k + 3k -4r +4r -2m -2m )  ( 3k - 3k -4r-4r -2m +2m)

= (6k - 4m) ( -8r)

= -48kr + 32mr

= 32mr - 48kr  

OR  16r (2m - 3k)                            [ If we take 16r common]

Hence, proved.

Hope you got that.

Thank You.