Question 1134148
{{{x^3+x-6=0 }}}

=>  the highest degree of its monomial (individual term) is {{{3}}}, so this function will have  one to three roots


use the Newton-Raphson method:

{{{x[n+1]}}}={{{x[n]}}}-{{{f(x[n])}}}/{{{f}}}'{{{(x[n])}}}


{{{f}}}'{{{(x[n])=3x^2+1}}}

let  {{{x[0]=5}}}

compute {{{x[n+1]}}} until \Delta \:{{{x[n+1] < 0.000001}}}

{{{f(x[0])=5^3+5-6=124}}}

{{{f}}}'{{{(x[0])=3*5^2+1=76}}}

{{{x[1]=5-124/76}}}

{{{x[1]=5-124/76}}}

{{{x[1]=3.36842}}}

 &Delta;{{{x[1]=5-3.36842}}}

 &Delta;{{{x[1]=highlight(1.6344)}}}=> one real root


Apply long division:

{{{(x^3+x-6)/(x-1.6344) =x^2+1.6344x+3.67115}}}


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


{{{x = (-1.6344 +- sqrt( 1.6344^2-4*1*3.67115 ))/(2*1) }}} 

{{{x = (-1.6344 +- sqrt( -12.01333664 ))/2 }}} 

{{{x = (-1.6344 +- 3.466*i)/2 }}} 


other roots:

{{{x = (-1.6344 + 3.466*i)/2 }}} =>{{{highlight(x[2] =-0.8172+1.733*i)}}}

{{{x = (-1.6344 - 3.466*i)/2 }}} =>{{{highlight(x[3] =-0.8172 -1.733*i)}}}