How do you find the real or imaginary solutions of the equation #x^3+64=0#?

1 Answer
Nov 27, 2016

Answer:

Use the sum of cubes identity to find the Real zero and a quadratic to solve and find the zeros are:

#4" "# and #" "2 +-2sqrt(3)i#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this later with #a=(x-2)# and #b=sqrt(-12) = 2sqrt(3)i#, but first...

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Note that #x^3# and #64 = 4^3# are both perfect squares, so the sub of cubes identity applies directly:

#x^3+64 = x^3+4^3#

#color(white)(x^3+64) = (x+4)(x^2-4x+16)#

We can factor the quadratic by completing the square:

#x^2-4x+16 = x^2-4x+4+12#

#color(white)(x^2-4x+16) = (x-2)^2-(2sqrt(3)i)^2#

#color(white)(x^2-4x+16) = ((x-2)-2sqrt(3)i)((x-2)+2sqrt(3)i)#

#color(white)(x^2-4x+16) = (x-2-2sqrt(3)i)(x-2+2sqrt(3)i)#

Hence the zeros of #x^3+64# are:

#4" "# and #" "2+-2sqrt(3)i#