Question 666718
{{{x^3-125 = 0 }}}

{{{x^3-5^3 = 0 }}}....use the difference of cubes rule {{{a^3 – b^3 = (a – b)(a^2 + ab + b^2)}}}

{{{(x-5)(x^2+5x+25) = 0 }}}

{{{(x-5)(x^2+5x+25) = 0 }}}

if {{{(x-5)= 0 }}}...=>...{{{x=5}}}......one real solution

if {{{(x^2+5x+25)= 0}}}...=>....use quadratic formula to fine to remaining solutions

{{{(x^2+5x+25)= 0}}}

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

{{{x = (-5 +- sqrt( 5^2-4*1*25 ))/(2*1) }}} 

{{{x = (-5 +- sqrt( 25-100 ))/2 }}} 

{{{x = (-5 +- sqrt( -75 ))/2 }}} 

{{{x = (-5 +- 8.66*i)/2 }}} 

solutions

{{{x = -5/2 + 8.66*i/2 }}}

{{{x = -2.5 + 4.33*i }}}

and

{{{x = -5/2 -8.66*i/2 }}}

{{{x = -2.5 -4.33*i }}}

these two are complex solutions


 {{{ graph( 600, 600, -10, 10, -200, 130, x^3-125) }}}