Question 340712
Find the zeros (or roots)

{{{x^3-8=0}}} express complex numbers in terms of i
<pre><b>
Factor the expression as the difference of cubes:

{{{(x-2)(x^2+2x+4)=0}}}

Set each factor = 0

Setting {{{x-2=0}}}
Solution: {{{x=2}}}

Setting the second factor = 0:

{{{x^2+2x+4=0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-(2) +- sqrt((2)^2-4*(1)*(4) ))/(2*(1)) }}} 

{{{x = (-2 +- sqrt(4-16))/2 }}}

{{{x = (-2 +- sqrt(-12))/2 }}}

{{{x = (-2 +- i*sqrt(12))/2 }}}

{{{x = (-2 +- i*sqrt(4*3))/2 }}}

{{{x = (-2 +- 2i*sqrt(3))/2 }}}

{{{x = (2(-1 +- i*sqrt(3)))/2 }}}

{{{x = (cross(2)(-1 +- i*sqrt(3)))/cross(2) }}}

{{{x = -1 +- i*sqrt(3) }}}

The three solutions are

{{{x=2}}}, {{{x = -1 + i*sqrt(3) }}}, {{{x = -1 - i*sqrt(3) }}}



Edwin</pre>