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Question Number 174177 by mathlove last updated on 26/Jul/22
x^3 +(1/x^3 )=1 then prove that (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3
$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1}\:\:\:\:\:{then}\:{prove}\:{that} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=\mathrm{3} \\ $$
Answered by mr W last updated on 26/Jul/22
(x+(1/x))^3 =x^3 +(1/x^3 )+3(x+(1/x)) P=x+(1/x) P^3 =3P+1 (x+(1/x))^2 =x^2 +(1/x^2 )+2 x^2 +(1/x^2 )=P^2 −2 (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=x^5 +(1/x^5 )+x+(1/x) x^5 +(1/x^5 )=P^2 −P−2 (x^5 +(1/x^5 ))^3 =P^6 −3P^5 −3P^4 +11P^3 +6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =(3P+1)^2 −3P^2 (3P+1)−3P(3P+1)+11(3P+1)+6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =9P^2 +6P+1−9P^3 −3P^2 −9P^2 −3P+33P+11+6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =−9P^3 +3P^2 +24P+4 (x^5 +(1/x^5 ))^3 =−9(3P+1)+3P^2 +24P+4 (x^5 +(1/x^5 ))^3 =3P^2 −3P−5 (x^5 +(1/x^5 ))^3 −1=3P^2 −3P−6 (x^5 +(1/x^5 ))^3 −1=3(P^2 −P−2)=3(x^5 +(1/x^5 )) ⇒(((x^5 +(1/x^5 ))^3 −1)/((x^5 +(1/x^5 ))))=3 ✓
$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${P}={x}+\frac{\mathrm{1}}{{x}} \\ $$$${P}^{\mathrm{3}} =\mathrm{3}{P}+\mathrm{1} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={P}^{\mathrm{2}} −\mathrm{2} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }={P}^{\mathrm{2}} −{P}−\mathrm{2} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} ={P}^{\mathrm{6}} −\mathrm{3}{P}^{\mathrm{5}} −\mathrm{3}{P}^{\mathrm{4}} +\mathrm{11}{P}^{\mathrm{3}} +\mathrm{6}{P}^{\mathrm{2}} −\mathrm{12}{P}−\mathrm{8} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =\left(\mathrm{3}{P}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}{P}^{\mathrm{2}} \left(\mathrm{3}{P}+\mathrm{1}\right)−\mathrm{3}{P}\left(\mathrm{3}{P}+\mathrm{1}\right)+\mathrm{11}\left(\mathrm{3}{P}+\mathrm{1}\right)+\mathrm{6}{P}^{\mathrm{2}} −\mathrm{12}{P}−\mathrm{8} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =\mathrm{9}{P}^{\mathrm{2}} +\mathrm{6}{P}+\mathrm{1}−\mathrm{9}{P}^{\mathrm{3}} −\mathrm{3}{P}^{\mathrm{2}} −\mathrm{9}{P}^{\mathrm{2}} −\mathrm{3}{P}+\mathrm{33}{P}+\mathrm{11}+\mathrm{6}{P}^{\mathrm{2}} −\mathrm{12}{P}−\mathrm{8} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =−\mathrm{9}{P}^{\mathrm{3}} +\mathrm{3}{P}^{\mathrm{2}} +\mathrm{24}{P}+\mathrm{4} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =−\mathrm{9}\left(\mathrm{3}{P}+\mathrm{1}\right)+\mathrm{3}{P}^{\mathrm{2}} +\mathrm{24}{P}+\mathrm{4} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =\mathrm{3}{P}^{\mathrm{2}} −\mathrm{3}{P}−\mathrm{5} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3}{P}^{\mathrm{2}} −\mathrm{3}{P}−\mathrm{6} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3}\left({P}^{\mathrm{2}} −{P}−\mathrm{2}\right)=\mathrm{3}\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right) \\ $$$$\Rightarrow\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)}=\mathrm{3}\:\checkmark \\ $$
Commented by mathlove last updated on 26/Jul/22
thanks mr W
$${thanks}\:{mr}\:{W} \\ $$
Answered by chengulapetrom last updated on 26/Jul/22
(x+(1/x))^5 =x^5 +5x^4 ((1/x))+10x^3 ((1/x))^2 +10x^2 ((1/x))^3 +5x((1/x))^4 +((1/x))^5 =x^5 +(1/x^5 )+5(x^3 +(1/x^3 ))+10(x+(1/x)) =x^5 +(1/x^5 )+10(x+(1/x))+5 (x+(1/x))^5 =x^5 +(1/x^5 )+10(x+(1/x))+5 (x+(1/x))^3 =x^3 +3(x+(1/x))+(1/x^3 ) (x+(1/x))^3 =3(x+(1/x))+1 (x+(1/x))^3 −3(x+(1/x))−1=0 (x+(1/x))^3 −1=3(x+(1/x)) (((x+(1/x))^3 −1)/((x+(1/x))))=3 similarly (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3
$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} ={x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} \left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{10}{x}^{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\mathrm{5}{x}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{4}} +\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} \\ $$$$={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{5}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)+\mathrm{10}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{10}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{5} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} ={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{10}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{5} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{1} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)}=\mathrm{3} \\ $$$${similarly}\:\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=\mathrm{3} \\ $$
Commented by chengulapetrom last updated on 26/Jul/22
Mr W am I correct
$${Mr}\:{W}\:{am}\:{I}\:{correct} \\ $$
Commented by mr W last updated on 27/Jul/22
i don′t think it′s “similary” that (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3 and (((x+(1/x))^3 −1)/((x+(1/x))))=3. you must show it. or maybe i didn′t understand you correctly. in fact (((x^3 +(1/x^3 ))^3 −1)/(x^3 +(1/x^3 )))≠3 and (((x^4 +(1/x^4 ))^3 −1)/(x^4 +(1/x^4 )))≠3 etc.
$${i}\:{don}'{t}\:{think}\:{it}'{s}\:“{similary}''\:{that} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=\mathrm{3}\:{and}\:\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)}=\mathrm{3}. \\ $$$${you}\:{must}\:{show}\:{it}.\:{or}\:{maybe}\:{i}\:{didn}'{t}\: \\ $$$${understand}\:{you}\:{correctly}. \\ $$$${in}\:{fact}\:\frac{\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}\neq\mathrm{3}\:{and} \\ $$$$\frac{\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}\neq\mathrm{3}\:{etc}. \\ $$
Answered by mr W last updated on 26/Jul/22
let P_n =x^n +(1/x^n ) P_1 =e_1 =p P_2 =e_1 P_1 −2e_2 =p^2 −2 P_n =e_1 P_(n−1) −e_2 P_(n−2) P_3 =e_1 P_2 −P_1 1=p(p^2 −2)−p p^3 −3p−1=0 P_4 =pP_3 −P_2 =p−p^2 +2=−p^2 +p+2 P_5 =pP_4 −P_3 =(−p^2 +p+2)p−1=−p^3 +p^2 +2p−1 P_5 =−3p−1+p^2 +2p−1=p^2 −p−2 ......
$${let}\:{P}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${P}_{\mathrm{1}} ={e}_{\mathrm{1}} ={p} \\ $$$${P}_{\mathrm{2}} ={e}_{\mathrm{1}} {P}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{2} \\ $$$${P}_{{n}} ={e}_{\mathrm{1}} {P}_{{n}−\mathrm{1}} −{e}_{\mathrm{2}} {P}_{{n}−\mathrm{2}} \\ $$$${P}_{\mathrm{3}} ={e}_{\mathrm{1}} {P}_{\mathrm{2}} −{P}_{\mathrm{1}} \\ $$$$\mathrm{1}={p}\left({p}^{\mathrm{2}} −\mathrm{2}\right)−{p} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{1}=\mathrm{0} \\ $$$${P}_{\mathrm{4}} ={pP}_{\mathrm{3}} −{P}_{\mathrm{2}} ={p}−{p}^{\mathrm{2}} +\mathrm{2}=−{p}^{\mathrm{2}} +{p}+\mathrm{2} \\ $$$${P}_{\mathrm{5}} ={pP}_{\mathrm{4}} −{P}_{\mathrm{3}} =\left(−{p}^{\mathrm{2}} +{p}+\mathrm{2}\right){p}−\mathrm{1}=−{p}^{\mathrm{3}} +{p}^{\mathrm{2}} +\mathrm{2}{p}−\mathrm{1} \\ $$$${P}_{\mathrm{5}} =−\mathrm{3}{p}−\mathrm{1}+{p}^{\mathrm{2}} +\mathrm{2}{p}−\mathrm{1}={p}^{\mathrm{2}} −{p}−\mathrm{2} \\ $$$$…… \\ $$
Answered by Rasheed.Sindhi last updated on 27/Jul/22
x^6 =x^3 −1 x^5 =((x^3 −1)/x) ▶x^5 +(1/x^5 )=((x^3 −1)/x)+(x/(x^3 −1)) =(((x^3 −1)^2 +x^2 )/(x(x^3 −1)))=((x^6 −2x^3 +1+x^2 )/(x(x^3 −1))) =((x^3 −1−2x^3 +1+x^2 )/(x(x^3 −1))) =((−x^3 +x^2 )/(x(x^3 −1)))=((−x(x−1))/((x−1)(x^2 +x+1))) =((−1)/((x^2 +x+1)/x))=((−1)/(x+(1/x)_() +1))=−(1/(p+1)) ▶(((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=(((−(1/(p+1)))^3 −1)/(−(1/(p+1)))) =((((1/(p+1)))^3 +1)/(1/(p+1))) =((((1/(p+1))+1)( ((1/(p+1)))^2 −((1/(p+1)))+1))/(1/(p+1))) =(((p+2)/(p+1)))(((1−(p+1)+(p+1)^2 )/((p+1)^2 )))(p+1) =(((p+2)(1−p−1+p^2 +2p+1))/((p+1)^2 )) =(((p+2)(p^2 +p+1))/((p+1)^2 )) =(((p−1+3)(p^2 +p+1))/((p+1)^2 )) =(((p−1)(p^2 +p+1)+3(p^2 +p+1))/((p+1)^2 )) =((p^3 −1+3p^2 +3p+3)/((p+1)^2 )) =((p^3 +3p^2 +3p+2)/((p+1)^2 ))−3+3 =((p^3 +3p^2 +3p+2−3p^2 −6p−3)/(p^2 +2p+1))+3 =((p^3 −3p−1)/(p^2 +2p+1))+3 =(((x+(1/x))^3 −3(x+(1/x))−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =((x^3 +(1/x^3 )+3(x+(1/x))−3(x+(1/x))−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =((1−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =3
$$ \\ $$$${x}^{\mathrm{6}} ={x}^{\mathrm{3}} −\mathrm{1} \\ $$$${x}^{\mathrm{5}} =\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}} \\ $$$$ \\ $$$$\blacktriangleright{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}}+\frac{{x}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\frac{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{3}} −\mathrm{1}\right)}=\frac{{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}+{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{3}} −\mathrm{1}\right)} \\ $$$$\:\:\:\:\:=\frac{{x}^{\mathrm{3}} −\cancel{\mathrm{1}}−\mathrm{2}{x}^{\mathrm{3}} +\cancel{\mathrm{1}}+{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{3}} −\mathrm{1}\right)} \\ $$$$\:\:\:\:\:=\frac{−{x}^{\mathrm{3}} +{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{3}} −\mathrm{1}\right)}=\frac{−{x}\cancel{\left({x}−\mathrm{1}\right)}}{\cancel{\left({x}−\mathrm{1}\right)}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$\:\:=\frac{−\mathrm{1}}{\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}}}=\frac{−\mathrm{1}}{\underset{} {\underbrace{{x}+\frac{\mathrm{1}}{{x}}}}+\mathrm{1}}=−\frac{\mathrm{1}}{{p}+\mathrm{1}} \\ $$$$ \\ $$$$\blacktriangleright\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=\frac{\left(−\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)^{\mathrm{3}} −\mathrm{1}}{−\frac{\mathrm{1}}{{p}+\mathrm{1}}} \\ $$$$=\frac{\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)^{\mathrm{3}} +\mathrm{1}}{\frac{\mathrm{1}}{{p}+\mathrm{1}}} \\ $$$$=\frac{\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}+\mathrm{1}\right)\left(\:\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)+\mathrm{1}\right)}{\frac{\mathrm{1}}{{p}+\mathrm{1}}} \\ $$$$=\left(\frac{{p}+\mathrm{2}}{\cancel{{p}+\mathrm{1}}}\right)\left(\frac{\mathrm{1}−\left({p}+\mathrm{1}\right)+\left({p}+\mathrm{1}\right)^{\mathrm{2}} }{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }\right)\cancel{\left({p}+\mathrm{1}\right)} \\ $$$$=\frac{\left({p}+\mathrm{2}\right)\left(\cancel{\mathrm{1}}−{p}−\cancel{\mathrm{1}}+{p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{1}\right)}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({p}+\mathrm{2}\right)\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({p}−\mathrm{1}+\mathrm{3}\right)\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)+\mathrm{3}\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{p}^{\mathrm{3}} −\mathrm{1}+\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{3}}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{2}}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{3}+\mathrm{3} \\ $$$$=\frac{{p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{2}−\mathrm{3}{p}^{\mathrm{2}} −\mathrm{6}{p}−\mathrm{3}}{{p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{1}}+\mathrm{3} \\ $$$$=\frac{{p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{1}}+\mathrm{3} \\ $$$$=\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}}+\mathrm{3} \\ $$$$=\frac{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\cancel{\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)}−\cancel{\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)}−\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}}+\mathrm{3} \\ $$$$=\frac{\mathrm{1}−\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}}+\mathrm{3} \\ $$$$=\mathrm{3} \\ $$$$ \\ $$

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