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

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

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Answer 1 · 2016-01-20T02:00:11

$x = \pm 2, - 1 \pm \sqrt{3} i, 1 \pm \sqrt{3} i$ $x=+-2,-1+-sqrt3i,1+-sqrt3i$

Explanation:

Knowing the following factoring techniques is imperative:

Difference of squares: $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$
Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$
Difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

$x^{6} - 64 = 0$ $x^6-64=0$

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

$(x^{3} + 8) (x^{3} - 8) = 0$ $color(red)((x^3+8))color(green)((x^3-8))=0$

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

$(x + 2) (x^{2} - 2 x + 4) (x - 2) (x^{2} + 2 x + 4) = 0$ $color(red)((x+2)(x^2-2x+4))color(green)((x-2)(x^2+2x+4))=0$

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

$x + 2 = 0 \Rightarrow x = - 2$ $x+2=0=>color(blue)(x=-2$
$x - 2 = 0 \Rightarrow x = 2$ $x-2=0=>color(blue)(x=2$

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

Solving $x^{2} - 2 x + 4 = 0$ $x^2-2x+4=0$ :

$x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm 2 \sqrt{3} i}{2} = 1 \pm \sqrt{3} i$ $color(blue)(x)=(2+-sqrt(4-16))/2=(2+-2sqrt3i)/2color(blue)(=1+-sqrt3i$

Solving $x^{2} + 2 x + 4 = 0$ $x^2+2x+4=0$ :

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2} = \frac{- 2 \pm 2 \sqrt{3} i}{2} = - 1 \pm \sqrt{3} i$ $color(blue)(x)=(-2+-sqrt(4-16))/2=(-2+-2sqrt3i)/2color(blue)(=-1+-sqrt3i$

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

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Explanation:

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How do you find all the real and complex roots of x^6-64=0x6−64=0x^6-64=0?

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