Question

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

Answer 1

`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)`

Answer 2

`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)`