X cubed - y cubed factor completely

Question

asked Oct 11, 2020 212k views

1 Answer

Kaaveh Mohamedi · Answer 1 · 2020-10-17T04:21:07+0000

ANSWER

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

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:

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