How do you factor #x^6-27#?

2 Answers
Aug 11, 2016

Answer:

#x^6 -27 = (x^2 -3)(x^4 + 3x^2 +9)#

Explanation:

The difference of two perfect cubes has a pattern which just has to be applied once you recognise that these terms are both cubes.

#x^3 - y^3 = (x-y)(x^2 + xy + y^2)#

#x^6 -27 = (x^2 -3)(x^4 + 3x^2 +9)#

First bracket: # (root3(x) " same sign " root3(y))rArr (x-y)#

Form the second bracket from the first.

#" square the first term " " change sign" " product of the terms " "PLUS " "square the second term ")#

#(x^2 + xy + y^2)#

Aug 11, 2016

Answer:

#x^6-27=color(green)((x-sqrt(3))(x+sqrt(3))(x^4+3x^2+9))#

Explanation:

Temporarily replacing #x^2# with #k#, we have
#color(white)("XXX")x^6-27=k^3-3^3#
and using the "difference of cubes"
#color(white)("XXXXXXX")=(k-3)(k^2+3k+9)#

Note by checking the discriminant we can see that #(k^2+3k+9)# has no real roots (and thus no real factors).

However since #(k-3)=(x^2-3)# we can factor, using the "difference of squares" to get
#color(white)("XXXXXXX")=(x-sqrt(3))(x+sqrt(3))(k^2+3k+9)#
or
#color(white)("XXXXXXX")=(x-sqrt(3))(x+sqrt(3))(x^4+3x^2+9)#