Question

How do you find all the real and complex roots of `x^6-64=0`?

Answer 1

`x=+-2,-1+-sqrt3i,1+-sqrt3i`

Explanation:

Knowing the following factoring techniques is imperative:

• Difference of squares: `a^2-b^2=(a+b)(a-b)`
• Sum of cubes: `a^3+b^3=(a+b)(a^2-ab+b^2)`
• Difference of cubes: `a^3-b^3=(a-b)(a^2+ab+b^2)`

`x^6-64=0`

Apply difference of squares: `x^6=(x^3)^2,64=8^2`.

`color(red)((x^3+8))color(green)((x^3-8))=0`

Use both sum & difference of cubes: `x^3=(x)^3,8=2^3`.

`color(red)((x+2)(x^2-2x+4))color(green)((x-2)(x^2+2x+4))=0`

From here, set each portion of the product equal to `0`. The linear factors `x+2` and `x-2` are easiest:

`x+2=0=>color(blue)(x=-2`
`x-2=0=>color(blue)(x=2`

The following two quadratic factors can be solved via completing the square or using the quadratic formula.

Solving `x^2-2x+4=0`:

`color(blue)(x)=(2+-sqrt(4-16))/2=(2+-2sqrt3i)/2color(blue)(=1+-sqrt3i`

Solving `x^2+2x+4=0`:

`color(blue)(x)=(-2+-sqrt(4-16))/2=(-2+-2sqrt3i)/2color(blue)(=-1+-sqrt3i`