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

Jan 20, 2016

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

Explanation:

Knowing the following factoring techniques is imperative:

• Difference of squares: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
• Sum of cubes: ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
• Difference of cubes: ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

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

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

$\textcolor{red}{\left({x}^{3} + 8\right)} \textcolor{g r e e n}{\left({x}^{3} - 8\right)} = 0$

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

$\textcolor{red}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)} \textcolor{g r e e n}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)} = 0$

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

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

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

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