Question

X cubed - y cubed factor completely​

Answer 1

`{x}^(3) - {y}^(3) = (x - y)[ {x}^(2) + xy + {y}^(2)]`

Step-by-step explanation

We want to factor:

`{x}^(3) - {y}^(3)`

completely.

Recall from binomial theorem that:

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

We make x³-y³ the subject to get:

`{x}^(3) - {y}^(3) = ( {x - y)}^(3) + 3 {x}^(2) y - \:3x {y}^(2)`

We now factor the right hand side to get;

`{x}^(3) - {y}^(3) = ( {x - y)}^(3) + 3 {x} y(x - y)`

We factor further to get,

`{x}^(3) - {y}^(3) = (x - y)[( {x - y)}^(2) + 3 {x} y]`

`{x}^(3) - {y}^(3) = (x - y)[ {x}^(2) - 2xy + {y}^(2) + 3 {x} y]`

This finally simplifies to:

`{x}^(3) - {y}^(3) = (x - y)[ {x}^(2) + xy + {y}^(2)]`