



So when
we have something that is made up of two different terms



that
have been multiplied times themselves three times, 


we
have the difference of two cubes. 





So
X ^{3}  1 factors like this ... 





X^{3} 
1 = (X  1)(X^{2} + X + 1) 





In
general, if you have a ^{3}X
^{3}  b
^{3}, it
factors to 





(a^{3}X^{3}
 b^{3}) = (a  b)(a^{2}X^{2} + abX + b^{2}) 





Examples: 


X^{3}
 8 = 
(X  2)(X^{2}
+ 2X + 4) 8 =
2^{3} 


27X^{3}
 1 = 
(3X  1)(9X^{2}
+ 3X + 1) 27 = 3^{3} 





Like
with the squares, the numbers 


don't
have to be cubes of integers. 





We
can even write X  1 


as
the difference of two cubes. 











We
couldn't factor X ^{2} + 1, can we factor X
^{3} + 1? 





Yes
we can!!! 





It
works out almost like the difference of two cubes. 





The
only changes are that the sign in the first factor is positive 


and
the middle sign in the second factor is a negative. 


So: 


X^{3} +
1 = (X + 1)(X^{2}  X + 1) 





Using
our friends "a" and "b." 


We
can say ... 





(a^{3}X^{3}
+ b^{3}) = (a^{2}X^{2}  abX + b^{2}) 





This
one is called factoring the sum of two cubes. 





Study
these two different factoring plans for a moment or two. 


The
only difference is where the negative sign goes ... 





(a^{3}X^{3}
 b^{3}) = (a^{2}X^{2} + abX + b^{2}) 


(a^{3}X^{3}
+ b^{3}) = (a^{2}X^{2}  abX + b^{2}) 





copyright 2005 Bruce Kirkpatrick 
