Question

How do you use pascals triangle to expand `(2x-3y)^3`?

Answer 1

Pascal's triangle up to the fifth term is:


You construct this by remembering that the first two rows are all `1`. Then, you add the `1, 1` to get the `2` right below and between the two `1`s from `1, 1`. Basically you add number pairs together and put the result in between and below the number pair. `3 + 3 = 6`, `4 + 6 = 10`, etc.

Starting on the first row, the order is `0`, and as you go down, it increases in integer steps. You can see that the third-order term is therefore `1, 3, 3, 1`. To use it, you have:

`(a + b)^3 = 1a^3b^0 + 3a^2b^1 + 3a^1b^2 + 1a^0b^3`
`(a - b)^3 = 1a^3b^0 - 3a^2b^1 + 3a^1b^2 - 1a^0b^3`

Yours obviously has a subtraction sign; thus, you have, with `a = 2x` and `b = 3y`:

`(2x - 3y)^3`
`= color(blue)(1)(2x)^color(green)(3)(3y)^color(green)(0) - color(blue)(3)(2x)^color(green)(2)(3y)^color(green)(1) + color(blue)(3)(2x)^color(green)(1)(3y)^color(green)(2) - color(blue)(1)(2x)^color(green)(0)(3y)^color(green)(3)`

The triangle coefficients are highlighted in blue. Notice how since it's a cube of a difference, the signs alternate.

Also, notice in green that the order of the first term `(2x)` starts at `3` and ends at `0`, while `(3y)` does the complete opposite, starting at `0` and ending at `3`.

Simplifying it:

`= (2x)^3 - 3(2x)^2(3y) + 3(2x)(3y)^2 - (3y)^3`

`= 8x^3 - (12x^2)(3y) + (6x)(9y^2) - (27y^3)`

`= color(blue)(8x^3 - 36x^2y + 54xy^2 - 27y^3)`