Expand the following
`(i) (3a-2b)^(3) (ii) ((1)/(x)+(y)/(3))^(3)` (iii) `(4-(1)/(3x))^(2)`

Explanation

(i) Putting `(1)/(x)=a and (y)/(3)=b`, we get
`((1)/(x)+(y)/(3))^3=(a+b)^3`
` =a^3+b^3+3ab(a+b)`
`=((1)/(x))^3+((y)/(3))^3+3xx(1)/(x)xx(y)/(3)xx((1)/((x)+(y)/(3)))`
`=(1)/(x^3)+(y^3)/(27)+(y)/(x)((1)/(x)+(y)/(3))`
`=(1)/(x^3)+(y^3)/(27)+(y)/(x^2)+(y^2)/(3x)=(1)/(x^3)+(y)/(x^2)+(y^2)/(3x)+(y^3)/(27)`.
(ii) Putting `4=a and (1)/(3x)=b`, we get
`(4-(1)/(3x))^2=(a-b)^3`
`a^3-b^3-3ab(a-b)`
`=(4)^3-((1)/(3x))^3-3xx4xx(1)/(3x)xx(4-(1)/(3x))`
`=64-(1)/(27x^3)-(16)/(x)+(4)/(3x^2)=64-(16)/(x)+(4)/(3x^2)-(1)/(27x^3)`