Question

How do you factor `x^6 + 8`?

Answer 1

You express `8` as a power of `2`.

`8 = underbrace(2 * 2 * 2)_(color(blue)("3 times")) = 2^(color(blue)(3))`

Now your expression becomes

`E = x^(6) + 2^(3)`

You can express `x^(6)` as a cube by using

`x^(6) = x^(2 * 3) = (x^(2))^(3)`

Now you have a sum of two cubes, which you can factor according to the formula

`a^3 + b^3 = (a + b)(a^2 -ab + b^2)`

So, your expression will become

`E = (x^2)^3 + 2^3 = (x^2 + 2)((x^2)^2 - (x^2)2 + 2^2)`

`E = (x^2 + 2)(x^4 - 2x^2 + 4)`

Answer 2

This polynomial can be seen as a sum of cubes.

Remember the rule of factorize a sum of cubes:

`a^3+b^3=(a+b)(a^2-ab+b^2)`.

`x^6+8=(x^2)^3+2^3=(x^2+2)(x^4-2x^2+4)=(1)`,

generally the second factor can't be further factored, but this happens only if it is a square polynomial, but... this is a 4th degree polynomial, so:

`x^4-2x^2+4=x^4+4x^2-6x^2 +4=x^4+4x^2+4-6x^2=`

`=(x^2+2)^2-(sqrt6x)^2=`

`=(x^2+2+sqrt6x)(x^2+2-sqrt6x)`,

and they can't be further factored.

Finally:

`(1)=(x^2+2)(x^2+2+sqrt6x)(x^2+2-sqrt6x)`.

Answer 3

`x^6` + 8 = (`x^2`+ 2)(`x^4`- 2`x^2` + 4)=(`x^2`+2)((`x^2`-1`)^2`+3)
Factors are: x + `sqrt(2)`i, x-`sqrt(2)`i, x-1+`sqrt(3)`i, x-1 - `sqrt(3)`i