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Question Number 202374 by Calculusboy last updated on 25/Dec/23

Commented by Frix last updated on 25/Dec/23

x^3 +px+q=0 Σ_(k=1) ^3 (((1+x_k )^2 )/((1−x_k )^2 )) =((3p^2 +2pq+3q^2 −10p−30q+3)/(p+q+1)) Σ_(k=1) ^3 (((1−x_k )^2 )/((1+x_k )^2 )) =((3p^2 −2pq+3q^2 −10p+30q+3)/(p−q+1))

$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\:\frac{\left(\mathrm{1}+{x}_{{k}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}_{{k}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{3}{p}^{\mathrm{2}} +\mathrm{2}{pq}+\mathrm{3}{q}^{\mathrm{2}} −\mathrm{10}{p}−\mathrm{30}{q}+\mathrm{3}}{{p}+{q}+\mathrm{1}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\:\frac{\left(\mathrm{1}−{x}_{{k}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}_{{k}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{2}{pq}+\mathrm{3}{q}^{\mathrm{2}} −\mathrm{10}{p}+\mathrm{30}{q}+\mathrm{3}}{{p}−{q}+\mathrm{1}} \\ $$

Commented by Calculusboy last updated on 26/Dec/23

nice solution

$$\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$

Answered by aleks041103 last updated on 25/Dec/23

x^3 −x−1 x=y+1 ⇒x^3 −x−1=(1+y)^3 −y−2= =y^3 +3y^2 +3y+1−y−2= =y^3 +3y^2 +2y−1=(y−(α−1))(y−(β−1))(y−(γ−1))= =(y−a)(y−b)(y−c) ⇒ { ((abc=1)),((ab+bc+ac=2)),((a+b+c=−3)) :} (((1+α)/(1−α)))^2 =(−(((1−α)−2)/(1−α)))^2 =(1+(2/(α−1)))^2 = =1+(4/(α−1))+(4/((α−1)^2 ))=1+(4/a)+(4/a^2 ) Σ(((1+α)/(1−α)))^2 =3+4Σ(1/a)+4Σ(1/a^2 ) Σ(1/a)=((ab+bc+ac)/(abc))=(2/1)=2 Σ(1/a^2 )=((a^2 b^2 +b^2 c^2 +a^2 c^2 )/((abc)^2 ))=(ab)^2 +(bc)^2 +(ac)^2 = =(ab+bc+ac)^2 −2(a^2 bc+ab^2 c+abc^2 )= =(ab+bc+ac)^2 −2abc(a+b+c)= =2^2 −2(−3)=10 ⇒Σ(((1+α)/(1−α)))^2 =3+4.2+4.10=51 ⇒Σ_(α,β,γ) (((1+x)/(1−x)))^2 =51, x^3 −x−1=0, ∀x∈{α,β,γ}

$${x}^{\mathrm{3}} −{x}−\mathrm{1} \\ $$$${x}={y}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{x}−\mathrm{1}=\left(\mathrm{1}+{y}\right)^{\mathrm{3}} −{y}−\mathrm{2}= \\ $$$$={y}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{1}−{y}−\mathrm{2}= \\ $$$$={y}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{1}=\left({y}−\left(\alpha−\mathrm{1}\right)\right)\left({y}−\left(\beta−\mathrm{1}\right)\right)\left({y}−\left(\gamma−\mathrm{1}\right)\right)= \\ $$$$=\left({y}−{a}\right)\left({y}−{b}\right)\left({y}−{c}\right) \\ $$$$\Rightarrow\begin{cases}{{abc}=\mathrm{1}}\\{{ab}+{bc}+{ac}=\mathrm{2}}\\{{a}+{b}+{c}=−\mathrm{3}}\end{cases} \\ $$$$ \\ $$$$\left(\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\right)^{\mathrm{2}} =\left(−\frac{\left(\mathrm{1}−\alpha\right)−\mathrm{2}}{\mathrm{1}−\alpha}\right)^{\mathrm{2}} =\left(\mathrm{1}+\frac{\mathrm{2}}{\alpha−\mathrm{1}}\right)^{\mathrm{2}} = \\ $$$$=\mathrm{1}+\frac{\mathrm{4}}{\alpha−\mathrm{1}}+\frac{\mathrm{4}}{\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{4}}{{a}}+\frac{\mathrm{4}}{{a}^{\mathrm{2}} } \\ $$$$\Sigma\left(\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\right)^{\mathrm{2}} =\mathrm{3}+\mathrm{4}\Sigma\frac{\mathrm{1}}{{a}}+\mathrm{4}\Sigma\frac{\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$$\Sigma\frac{\mathrm{1}}{{a}}=\frac{{ab}+{bc}+{ac}}{{abc}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2} \\ $$$$\Sigma\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{\left({abc}\right)^{\mathrm{2}} }=\left({ab}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} +\left({ac}\right)^{\mathrm{2}} = \\ $$$$=\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{2}} {bc}+{ab}^{\mathrm{2}} {c}+{abc}^{\mathrm{2}} \right)= \\ $$$$=\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} −\mathrm{2}{abc}\left({a}+{b}+{c}\right)= \\ $$$$=\mathrm{2}^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{3}\right)=\mathrm{10} \\ $$$$\Rightarrow\Sigma\left(\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\right)^{\mathrm{2}} =\mathrm{3}+\mathrm{4}.\mathrm{2}+\mathrm{4}.\mathrm{10}=\mathrm{51} \\ $$$$\Rightarrow\underset{\alpha,\beta,\gamma} {\sum}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} =\mathrm{51},\:{x}^{\mathrm{3}} −{x}−\mathrm{1}=\mathrm{0},\:\forall{x}\in\left\{\alpha,\beta,\gamma\right\} \\ $$

Commented by Calculusboy last updated on 25/Dec/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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