Question 1178729


{{{x^3 - 2x + 1 = 0}}}..........factor

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

one root is {{{r=1}}}


use quadratic formula to find other two roots for

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


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


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


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


{{{x=(-1+-sqrt(5))/2}}}


=> {{{s=(-1+sqrt(5))/2}}} =>{{{s=-1/2+sqrt(5)/2}}}

and {{{t=(-1-sqrt(5))/2}}}=>{{{t=-1/2-sqrt(5)/2}}}


then

{{{r^3 - s- t=1^3 - (-1/2+sqrt(5)/2)- (-1/2-sqrt(5)/2)}}}

{{{r^3 - s- t=1+1/2-sqrt(5)/2)+1/2+sqrt(5)/2}}}

{{{r^3 - s- t=1+1/2+1/2}}}

{{{r^3 - s- t=1+1}}}

{{{r^3 - s- t=2}}}