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Question Number 205073 by mr W last updated on 07/Mar/24
if a, b, c are the roots of f(x)=x^3 −2024x^2 +2024x+2024 find (1/(1−a^2 ))+(1/(1−b^2 ))+(1/(1−c^2 ))=?
$${if}\:{a},\:{b},\:{c}\:{are}\:{the}\:{roots}\:{of} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2024}{x}^{\mathrm{2}} +\mathrm{2024}{x}+\mathrm{2024} \\ $$$${find}\:\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{c}^{\mathrm{2}} }=? \\ $$
Answered by cortano12 last updated on 07/Mar/24
= (a^2 /(1−a^2 )) +(b^2 /(1−b^2 )) +(c^2 /(1−c^2 )) + 3 = ((a^2 +b^2 −2(ab)^2 )/((1−a^2 )(1−b^2 ))) +(c^2 /(1−c^2 )) + 3 =((a^2 +b^2 +c^2 +3(abc)^2 −2((ab)^2 +(ac)^2 +(bc)^2 )/((1−a^2 )(1−b^2 )(1−c^2 )))+3
$$\:=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\:+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{1}−\mathrm{b}^{\mathrm{2}} }\:+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:+\:\mathrm{3}\: \\ $$$$\:=\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{b}^{\mathrm{2}} \right)}\:+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:+\:\mathrm{3}\: \\ $$$$\:=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{abc}\right)^{\mathrm{2}} −\mathrm{2}\left(\left(\mathrm{ab}\right)^{\mathrm{2}} +\left(\mathrm{ac}\right)^{\mathrm{2}} +\left(\mathrm{bc}\right)^{\mathrm{2}} \right.}{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{c}^{\mathrm{2}} \right)}+\mathrm{3} \\ $$
Answered by Frix last updated on 07/Mar/24
y=(1/(1−x^2 )) ⇒ x=±(√((y−1)/y)) Inserting & transforming y^3 −((2027y^2 )/(2025))−((2021y)/(2025))−(1/(4100625))=0 ⇒ answer is ((2027)/(2025))
$${y}=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\:{x}=\pm\sqrt{\frac{{y}−\mathrm{1}}{{y}}} \\ $$$$\mathrm{Inserting}\:\&\:\mathrm{transforming} \\ $$$${y}^{\mathrm{3}} −\frac{\mathrm{2027}{y}^{\mathrm{2}} }{\mathrm{2025}}−\frac{\mathrm{2021}{y}}{\mathrm{2025}}−\frac{\mathrm{1}}{\mathrm{4100625}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{2027}}{\mathrm{2025}} \\ $$
Answered by A5T last updated on 07/Mar/24
WLOG ,let a=x+yi,b=x−yi,c=c where x,y,c∈R ?=(1/(1−(x+yi)^2 ))+(1/(1−(x−yi)^2 ))+(1/(1−c^2 )) ⇒2?=(1/(1−x−yi))+(1/(1−x+yi))+(1/(1−c))+(1/(1+x+yi))+(1/(1+x−yi))+(1/(1+c)) =((2(1−x))/((1−x)^2 +y^2 ))+(1/(1−c))+((2(1+x))/((1+x)^2 +y^2 ))+(1/(1+c)) =((2(1−c−x+xc)+1+x^2 −2x+y^2 )/(1+x^2 −2x+y^2 −c−cx^2 +2cx−cy^2 ))+((2(1+x+c+cx)+1+x^2 +2x+y^2 )/(1+x^2 +2x+y^2 +c+cx^2 +2cx+cy^2 )) =((3−2(c+2x)+x^2 +y^2 +2cx)/(1+x^2 +y^2 +2cx−(2x+c)−c(x^2 +y^2 )))+((3+x^2 +y^2 +2cx+2(2x+c))/(1+x^2 +y^2 +2cx+2x+c+c(x^2 +y^2 ))) =((3−(2024))/(2025))+((3×2025)/(2025))=((4054)/(2025))=2? ⇒?=((2027)/(2025)) [ab+bc+ca=2024⇒x^2 +y^2 +2cx=2024 abc=(x^2 +y^2 )c=−2024;a+b+c=2x+c=2024]
$${WLOG}\:,{let}\:{a}={x}+{yi},{b}={x}−{yi},{c}={c}\:{where}\:{x},{y},{c}\in\mathbb{R} \\ $$$$?=\frac{\mathrm{1}}{\mathrm{1}−\left({x}+{yi}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left({x}−{yi}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}?=\frac{\mathrm{1}}{\mathrm{1}−{x}−{yi}}+\frac{\mathrm{1}}{\mathrm{1}−{x}+{yi}}+\frac{\mathrm{1}}{\mathrm{1}−{c}}+\frac{\mathrm{1}}{\mathrm{1}+{x}+{yi}}+\frac{\mathrm{1}}{\mathrm{1}+{x}−{yi}}+\frac{\mathrm{1}}{\mathrm{1}+{c}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{c}}+\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−{c}−{x}+{xc}\right)+\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}+{y}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}+{y}^{\mathrm{2}} −{c}−{cx}^{\mathrm{2}} +\mathrm{2}{cx}−{cy}^{\mathrm{2}} }+\frac{\mathrm{2}\left(\mathrm{1}+{x}+{c}+{cx}\right)+\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{2}{x}+{y}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{2}{x}+{y}^{\mathrm{2}} +{c}+{cx}^{\mathrm{2}} +\mathrm{2}{cx}+{cy}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}−\mathrm{2}\left({c}+\mathrm{2}{x}\right)+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{cx}}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{cx}−\left(\mathrm{2}{x}+{c}\right)−{c}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}+\frac{\mathrm{3}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{cx}+\mathrm{2}\left(\mathrm{2}{x}+{c}\right)}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{cx}+\mathrm{2}{x}+{c}+{c}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{3}−\left(\mathrm{2024}\right)}{\mathrm{2025}}+\frac{\mathrm{3}×\mathrm{2025}}{\mathrm{2025}}=\frac{\mathrm{4054}}{\mathrm{2025}}=\mathrm{2}?\:\:\Rightarrow?=\frac{\mathrm{2027}}{\mathrm{2025}} \\ $$$$\left[{ab}+{bc}+{ca}=\mathrm{2024}\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{cx}=\mathrm{2024}\right. \\ $$$$\left.{abc}=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){c}=−\mathrm{2024};{a}+{b}+{c}=\mathrm{2}{x}+{c}=\mathrm{2024}\right] \\ $$
Answered by A5T last updated on 07/Mar/24
(1/(1−a))+(1/(1−b))+(1/(1−c))+(1/(1+a))+(1/(1+b))+(1/(1+c))=2? ((Σ(1−a)(1−b))/((1−a)(1−b)(1−c)))+((Σ(1+a)(1+b))/((1+a)(1+b)(1+c))) =((3−2(a+b+c)+ab+bc+ca)/(1−a−b−c+ab+ac+bc−abc))+((3+2(Σa)+Σab)/(1+Σa+Σab+abc)) =((3−2(2024)+2024)/(1−2024+2024+2024))+((3+2(2024)+2024)/(1+2024+2024−2024)) =((3−2024)/(2025))+((3×2025)/(2025))=((4054)/(2025))=2?⇒?=((2027)/(2025))
$$\frac{\mathrm{1}}{\mathrm{1}−{a}}+\frac{\mathrm{1}}{\mathrm{1}−{b}}+\frac{\mathrm{1}}{\mathrm{1}−{c}}+\frac{\mathrm{1}}{\mathrm{1}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{c}}=\mathrm{2}? \\ $$$$\frac{\Sigma\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)}+\frac{\Sigma\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)} \\ $$$$=\frac{\mathrm{3}−\mathrm{2}\left({a}+{b}+{c}\right)+{ab}+{bc}+{ca}}{\mathrm{1}−{a}−{b}−{c}+{ab}+{ac}+{bc}−{abc}}+\frac{\mathrm{3}+\mathrm{2}\left(\Sigma{a}\right)+\Sigma{ab}}{\mathrm{1}+\Sigma{a}+\Sigma{ab}+{abc}} \\ $$$$=\frac{\mathrm{3}−\mathrm{2}\left(\mathrm{2024}\right)+\mathrm{2024}}{\mathrm{1}−\mathrm{2024}+\mathrm{2024}+\mathrm{2024}}+\frac{\mathrm{3}+\mathrm{2}\left(\mathrm{2024}\right)+\mathrm{2024}}{\mathrm{1}+\mathrm{2024}+\mathrm{2024}−\mathrm{2024}} \\ $$$$=\frac{\mathrm{3}−\mathrm{2024}}{\mathrm{2025}}+\frac{\mathrm{3}×\mathrm{2025}}{\mathrm{2025}}=\frac{\mathrm{4054}}{\mathrm{2025}}=\mathrm{2}?\Rightarrow?=\frac{\mathrm{2027}}{\mathrm{2025}} \\ $$
Answered by mr W last updated on 07/Mar/24
(x−1+1)^3 −2024(x−1+1)^2 +2024(x−1+1)+2024=0 let t=x−1 (t+1)^3 −2024(t+1)^2 +2024(t+1)+2024=0 ((2025)/t^3 )−((2021)/t^2 )−((2021)/t)+1=0 ⇒Σ(1/(1−x))=−Σ(1/t)=−((2021)/(2025)) (x+1−1)^3 −2024(x+1−1)^2 +2024(x+1−1)+2024=0 let s=x+1 (s−1)^3 −2024(s−1)^2 +2024(s−1)+2024=0 s^3 −2027s^2 +6075s−2025=0 ((2025)/s^3 )−((6075)/s^2 )+((2027)/s)−1=0 ⇒Σ(1/(1+x))=Σ(1/s)=((6075)/(2025)) (1/(1−a^2 ))+(1/(1−b^2 ))+(1/(1−c^2 )) =(1/2)((1/(1−a))+(1/(1−b))+(1/(1−c))+(1/(1+a))+(1/(1+b))+(1/(1+c))) =(1/2)(Σ(1/(1−x))+Σ(1/(1+x))) =(1/2)(−((2021)/(2025))+((6075)/(2025)))=((2027)/(2025)) ✓
$$\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2024}\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2024}\left({x}−\mathrm{1}+\mathrm{1}\right)+\mathrm{2024}=\mathrm{0} \\ $$$${let}\:{t}={x}−\mathrm{1} \\ $$$$\left({t}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2024}\left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2024}\left({t}+\mathrm{1}\right)+\mathrm{2024}=\mathrm{0} \\ $$$$\frac{\mathrm{2025}}{{t}^{\mathrm{3}} }−\frac{\mathrm{2021}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2021}}{{t}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{\mathrm{1}−{x}}=−\Sigma\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{2021}}{\mathrm{2025}} \\ $$$$\left({x}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2024}\left({x}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2024}\left({x}+\mathrm{1}−\mathrm{1}\right)+\mathrm{2024}=\mathrm{0} \\ $$$${let}\:{s}={x}+\mathrm{1} \\ $$$$\left({s}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2024}\left({s}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2024}\left({s}−\mathrm{1}\right)+\mathrm{2024}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} −\mathrm{2027}{s}^{\mathrm{2}} +\mathrm{6075}{s}−\mathrm{2025}=\mathrm{0} \\ $$$$\frac{\mathrm{2025}}{{s}^{\mathrm{3}} }−\frac{\mathrm{6075}}{{s}^{\mathrm{2}} }+\frac{\mathrm{2027}}{{s}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{\mathrm{1}+{x}}=\Sigma\frac{\mathrm{1}}{{s}}=\frac{\mathrm{6075}}{\mathrm{2025}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{c}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}−{a}}+\frac{\mathrm{1}}{\mathrm{1}−{b}}+\frac{\mathrm{1}}{\mathrm{1}−{c}}+\frac{\mathrm{1}}{\mathrm{1}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{c}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\Sigma\frac{\mathrm{1}}{\mathrm{1}−{x}}+\Sigma\frac{\mathrm{1}}{\mathrm{1}+{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{2021}}{\mathrm{2025}}+\frac{\mathrm{6075}}{\mathrm{2025}}\right)=\frac{\mathrm{2027}}{\mathrm{2025}}\:\checkmark \\ $$
Answered by pi314 last updated on 08/Mar/24
((P′(x))/(P(x)))=Σ_(a,b,c) (1/(x−c));((3x^2 −4048x+2024)/(x^3 −2024x^2 +2024x+2024)) S=(1/2)((1/(1−a))+(1/(1−b))+(1/(1−c)))+(1/2)((1/(1+a))+(1/(1+b))+(1/(1+c))) S=(1/2)(((P′(1))/(P(1)))−((P′(−1))/(P(−1))))=(1/2)(((−2021)/(2025))+((6075)/(2025)))=((2027)/(2025))
$$\frac{{P}'\left({x}\right)}{{P}\left({x}\right)}=\underset{{a},{b},{c}} {\sum}\frac{\mathrm{1}}{{x}−{c}};\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4048}{x}+\mathrm{2024}}{{x}^{\mathrm{3}} −\mathrm{2024}{x}^{\mathrm{2}} +\mathrm{2024}{x}+\mathrm{2024}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}−{a}}+\frac{\mathrm{1}}{\mathrm{1}−{b}}+\frac{\mathrm{1}}{\mathrm{1}−{c}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{c}}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{P}'\left(\mathrm{1}\right)}{{P}\left(\mathrm{1}\right)}−\frac{{P}'\left(−\mathrm{1}\right)}{{P}\left(−\mathrm{1}\right)}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{2021}}{\mathrm{2025}}+\frac{\mathrm{6075}}{\mathrm{2025}}\right)=\frac{\mathrm{2027}}{\mathrm{2025}} \\ $$

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