Question 355697
It's the difference of two cubes and can be factored.
{{{8x^3-1=(2x-1)(4x^2+2x+1)=0}}}
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Real solution:
{{{2x-1=0}}}
{{{2x=1}}}
{{{highlight(x=1/2)}}}
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Complex solution:
{{{4x^2+2x+1=0}}}
Use the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-2+- sqrt( 2^2-4*4*1 ))/(2*4) }}} 
{{{x = (-2+- sqrt( 4-16 ))/8 }}} 
{{{x = (-2+- sqrt( -12 ))/8 }}}
{{{x = (-2+- 2*sqrt( -3 ))/8 }}}
{{{highlight(x = (-1+- sqrt( 3 )i)/4)}}} where {{{i=sqrt(-1)}}}