Menu Close

x-3-x-c-let-x-mt-1-t-m-3-t-3-mt-1-t-2-c-1-t-3-m-3-m-c-t-3-2m-3c-t-2-3c-m-t-c-0-t-3-At-2-Bt-C-0-let-AB-C-2m-3c-m-3c-c-m-3-m-c-m-3-2-c-m-2-8m-10c-0-




Question Number 187166 by ajfour last updated on 14/Feb/23
x^3 =x+c  let  x=((mt)/(1−t))  m^3 t^3 =mt(1−t)^2 +c(1−t)^3   ⇒  (m^3 −m−c)t^3 +(2m−3c)t^2        +(3c−m)t−c=0  t^3 +At^2 +Bt+C=0  let  AB=C  ⇒ (2m−3c)(m−3c)=c(m^3 −m−c)  ⇒ m^3 −(2/c)m^2 −8m+10c=0  ...
$${x}^{\mathrm{3}} ={x}+{c} \\ $$$${let}\:\:{x}=\frac{{mt}}{\mathrm{1}−{t}} \\ $$$${m}^{\mathrm{3}} {t}^{\mathrm{3}} ={mt}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left({m}^{\mathrm{3}} −{m}−{c}\right){t}^{\mathrm{3}} +\left(\mathrm{2}{m}−\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left(\mathrm{3}{c}−{m}\right){t}−{c}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +{At}^{\mathrm{2}} +{Bt}+{C}=\mathrm{0} \\ $$$${let}\:\:{AB}={C} \\ $$$$\Rightarrow\:\left(\mathrm{2}{m}−\mathrm{3}{c}\right)\left({m}−\mathrm{3}{c}\right)={c}\left({m}^{\mathrm{3}} −{m}−{c}\right) \\ $$$$\Rightarrow\:{m}^{\mathrm{3}} −\frac{\mathrm{2}}{{c}}{m}^{\mathrm{2}} −\mathrm{8}{m}+\mathrm{10}{c}=\mathrm{0} \\ $$$$… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *