Question

Expand the following:

`(3x-1/(3x))^3`

Answer 1

Using the identity:

(a – b)³ = a³ – 3ab(a – b) – b³

Here, let a = 3x, b = `1/(3x)`

`(3x-1/(3x))^3 = (3x)^3 - 3(3x)(1/(3x))(3x - 1/(3x))-(1/(3x))^3`

= `27x^3-3(1)(3x - 1/(3x)) - 1/(27x^3)`

= `27x^3 - 3(3x - 1/(3x)) - 1/(27x^3)`

= `27x^3 - (9x - 3/(x)) - 1/(27x^3)`

∴ `(3x - 1/(3x))^3 = 27x^3 - 9x+3/x - 1/(27x^3)`

Answer 2

Given: `(3x - 1/(3x))^3`

Stepwise calculation:

Use the binomial expansion formula for (a – b)³ = a³ – 3a²b + 3ab² – b³, where a = 3x and `b = 1/(3x)`.

1. Compute a³ = (3x)³ = 27x³.

2. Compute

`3a^2b = 3 xx (3x)^2 xx 1/(3x)`

`3a^2b = 3 xx 9x^2 xx 1/(3x)`

3a²b = 3 × 3x

3a²b = 9x. 

3. Compute

`3ab^2 = 3 xx (3x) xx (1/(3x))^2`

`3ab^2 = 3 xx 3x xx 1/(9x^2)`

`3ab^2 = 3 xx 3/(9x)`

`3ab^2 = 9/(9x)`

`3ab^2 = 1/x`.

4. Compute

`b^3 = (1/(3x))^3`

`b^3 = 1/(27x^3)`

Now putting it all together for (a – b)³:

`(3x - 1/(3x))^3 = 27x^3 - 9x + 1/x - 1/(27x^3)`