Menu Close

x-1-x-2-1-x-1-15-1-x-1-15-

Tinku Tara 0 Comments
Pin




Question Number 223483 by fantastic last updated on 26/Jul/25
(x−1)(x−2)=1 (x−1)^(15) −(1/((x−1)^(15) ))=??
$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{15}} −\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{15}} }=?? \\ $$
Answered by mr W last updated on 26/Jul/25
let t=x−1 t(t−1)=1 ⇒t−(1/t)=1 (t−(1/t))^3 =t^3 −(1/t^3 )−3(t−(1/t))=1^3 ⇒t^3 −(1/t^3 )=3+1^3 =4 ⇒(t^3 −(1/t^3 ))^2 =t^6 +(1/t^6 )−2=4^2 ⇒t^6 +(1/t^6 )=2+4^2 =18 ⇒(t^3 −(1/t^3 ))^3 =t^9 −(1/t^9 )−3(t^3 −(1/t^3 ))=4^3 ⇒t^9 −(1/t^9 )=3×4+4^3 =76 ⇒(t^9 −(1/t^9 ))(t^6 +(1/t^6 ))=76×18 ⇒t^(15) −(1/t^(15) )+t^3 −(1/t^3 )=76×18 ⇒t^(15) −(1/t^(15) )=76×18−4=1364 i.e. (x−1)^(15) −(1/((x−1)^(15) ))=1364 ✓
$${let}\:{t}={x}−\mathrm{1} \\ $$$${t}\left({t}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow{t}−\frac{\mathrm{1}}{{t}}=\mathrm{1} \\ $$$$\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} ={t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−\mathrm{3}\left({t}−\frac{\mathrm{1}}{{t}}\right)=\mathrm{1}^{\mathrm{3}} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }=\mathrm{3}+\mathrm{1}^{\mathrm{3}} =\mathrm{4} \\ $$$$\Rightarrow\left({t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right)^{\mathrm{2}} ={t}^{\mathrm{6}} +\frac{\mathrm{1}}{{t}^{\mathrm{6}} }−\mathrm{2}=\mathrm{4}^{\mathrm{2}} \\ $$$$\Rightarrow{t}^{\mathrm{6}} +\frac{\mathrm{1}}{{t}^{\mathrm{6}} }=\mathrm{2}+\mathrm{4}^{\mathrm{2}} =\mathrm{18} \\ $$$$\Rightarrow\left({t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right)^{\mathrm{3}} ={t}^{\mathrm{9}} −\frac{\mathrm{1}}{{t}^{\mathrm{9}} }−\mathrm{3}\left({t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right)=\mathrm{4}^{\mathrm{3}} \\ $$$$\Rightarrow{t}^{\mathrm{9}} −\frac{\mathrm{1}}{{t}^{\mathrm{9}} }=\mathrm{3}×\mathrm{4}+\mathrm{4}^{\mathrm{3}} =\mathrm{76} \\ $$$$\Rightarrow\left({t}^{\mathrm{9}} −\frac{\mathrm{1}}{{t}^{\mathrm{9}} }\right)\left({t}^{\mathrm{6}} +\frac{\mathrm{1}}{{t}^{\mathrm{6}} }\right)=\mathrm{76}×\mathrm{18} \\ $$$$\Rightarrow{t}^{\mathrm{15}} −\frac{\mathrm{1}}{{t}^{\mathrm{15}} }+{t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }=\mathrm{76}×\mathrm{18} \\ $$$$\Rightarrow{t}^{\mathrm{15}} −\frac{\mathrm{1}}{{t}^{\mathrm{15}} }=\mathrm{76}×\mathrm{18}−\mathrm{4}=\mathrm{1364} \\ $$$${i}.{e}.\:\left({x}−\mathrm{1}\right)^{\mathrm{15}} −\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{15}} }=\mathrm{1364}\:\checkmark \\ $$
Commented by fantastic last updated on 26/Jul/25
��

Pin

Leave a Reply

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