Question

X cubed - y cubed factor completely​

Answer 1

ANSWER

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

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:

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