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(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
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