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

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Answer 1 · 2016-01-20T02:00:11

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

Knowing the following factoring techniques is imperative:

$x^6-64=0$

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 $x^2+2x+4=0$ :

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

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