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

1 Answer
Jan 20, 2016

Answer:

#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#