Question

expand the following ​

Answer 1

Step-by

-step explanation:

hope you will understand this answer

Answer 2

(i) ( 3a/2 + b/4 - 2c ) ^2

let b/4 - 2c = x

(3a/2 + x)^2

• Using (a+b)^2 = a^2 + b^2 + 2ab

9a^2/4 + x^2 + 6ax/2

9a^2/4 + x^2 + 3ax

putting value of x

9a^2/4 + (b/4 -2c)^2 + 3a (b/4 - 2c)

• Using (a-b)^2 = a^2 + b^2 - 2ab

9a^2/4 + ( b^2/16 + 4c^2 - 4bc/4 ) + 3ab/4 - 6ac

9a^2/4 +  b^2/16 + 4c^2 - bc  + 3ab/4 - 6ac

OR (Another Way)

( 3a/2 + b/4 - 2c ) ^2

• using ( a + b - c )^2 = a^2 + b^2 + c^2 + 2ab – 2bc - 2ca

9a^2/4 + b^2/16 + 4c^2 + 6ab/8 - 4bc/4 - 12ca/2

9a^2/4 + b^2/16 + 4c^2 + 3ab/4 - bc - 6ac

(ii) (3/x - y/3)^3

• Using (a - b) ^3 = a^3 - 3a^2b + 3ab^2 - b^3

9/x^3 - 3 (3/x)^2 (y/3) + 3 (3/x)(y/3)^2 - y^3/9

9/x^3 - 27/x^2 (y/3) + 9/x(y^2/9) - y^3/9

9/x^3 - 9y/x^2 + y^2/x - y^3/9

Also see the attachment