# Thread: factorising - difference of squares

1. ## factorising - difference of squares

hi,

i have been asked to factorise the following

$l^8-256$

using the difference of squares $a^2-b^2=(a+b)(a-b)$ and by working from the lowest term $(l^2-2^2)$, i managed to get a solution but its was more down to guesswork than a systematic attempt to solve this.

$(l^4+16)(l^2+4)(l+2)(l-2)$

is there a way of dealing with these type of questions?

thanks

2. Originally Posted by sammy28
hi,

i have been asked to factorise the following

$l^8-265$

using the difference of squares $a^2-b^2=(a+b)(a-b)$ and by working from the lowest term $(l^2-2^2)$, i managed to get a solution but its was more down to guesswork than a systematic attempt to solve this.

$(l^4+16)(l^2+4)(l+2)(l-2)$

is there a way of dealing with these type of questions?

thanks
Hi sammy28,

Isn't that supposed to be $l^8-256$?

3. oops,

4. Originally Posted by masters
Hi sammy28,

Isn't that supposed to be $l^8-256$?
$l^8-256$ is the difference of two squares. Look at it this way:

$((l^4)^2-16^2)$

$(l^4-16)(l^4+16)$

$((l^2)^2-4^2)(l^4+16)$

$(l^2-4)(l^2+4)(l^4+16)$

$(l-2)(l+2)(l^2+4)(l^4+16)$

5. thanks masters