Does factoring by sum and difference of cube or sum and difference of squares yield the same answer Q) Factor: $s^6-1$
If I start by doing sum of difference of squares I get
$(s^3)^2 - (1)^2$
$(s^3-1)(s^3+1)$
$(s+1)(s^2-s+1)(s-1)(s^2+s+1)$
If I start by doing sum or difference of cubes I get
$S^6 - 1$
$(s^2)^3 - (1)^3$
$(s^2 - 1) (s^4+s^2 + 1)$
ANS: $(s-1)(s+1) (s^4+s^2 + 1)$
Both the answers are different, as a matter of fact I know that when I started by doing the sum or difference of square method, the answer I got was correct , but what am I doing wrong in the 2nd method then?

$s^4+s^2+1=(s^2-s+1)(s^2+s+1)$ so it is the same answer.
The way to complete the factoring is:
$s^4+s^2+1=s^4+2s^2+1-s^2=(s^2+1)^2-s^2=(s^2-s+1)(s^2+s+1)$
