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

1 Answer
Jan 20, 2016

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