Question

please solve but fast question no 45 a and b also

Answer 1

Here is your solution :

( a ).

= 4a² - 9b² - ( 2a - 3b )

= ( 2a )² - ( 3b )² - ( 2a - 3b )

Using identity,

=> ( a² - b² ) = ( a + b ) ( a - b )

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

Taking out ( 2a - 3b ) as common,

= ( 2a - 3b ) ( 2a + 3b - 1 )

( b ).

= a² + b² - 2( ab - ac + bc )

Using identity,

=> ( a² + b² ) = [ ( a - b )² + 2ab ]

= ( a - b )² + 2ab - 2( ab - ac + bc )

= ( a - b )² + 2ab - 2ab + 2ac - 2bc

= ( a - b )² + 2ac - 2bc

= ( a - b )² + 2c ( a - b )

Taking out ( a - b ) as common,

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

Proof of identities used in this question.

( i )

=> ( a² - b² ) = ( a + b ) ( a - b )

L.H.S = ( a² - b² )

R.H.S = ( a + b ) ( a - b )

= a( a - b ) + b( a - b )

= a² - ab + ab - b²

= a² - b² ( L.H.S )

Proved !!

( ii )

=> ( a² + b² ) = ( a - b )² + 2ab

Using identity,

=> ( a - b )² = a² + b² - 2ab

Adding ( 2ab ) to both sides,

=> ( a - b )² + 2ab = a² + b² - 2ab + 2ab

=> ( a - b )² + 2ab = a² + b²

=> -a² - b² = - ( a - b )² - 2ab

=> - ( a² + b² ) = - [ ( a - b )² + 2ab ]

•°• ( a² + b² ) = ( a - b )² + 2ab

Proof of identity used in proving this identity.

=> ( a - b )² = ( a² + b² - 2ab )

Now,

= ( a - b )² ( L.H.S )

= ( a - b ) ( a - b )

= a( a - b ) -b ( a - b )

= a² - ab - ab + b²

= a² - 2ab + b²

= a² + b² - 2ab ( R.H.S )

Proved !!