Question 934640
{{{f(x) = x^3 + 1}}}

set {{{f(x) =0}}} and use rule: {{{(a+b) (a^2-ab+b^2)}}}; in your case {{{a=x}}} and {{{b=1}}}


{{{0=(x+1)(x^2-x+1)}}}

solutions:


if {{{0=(x+1)}}}=> {{{x=-1}}}-real solution



if {{{0=(x^2-x+1)}}}-for this one, use quadratic formula:

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

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


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

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

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

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

complex solutions:

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

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


{{{ graph( 600, 600, -10, 10, -10, 10, x^3 + 1) }}}