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Question-78814

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Question Number 78814 by M±th+et£s last updated on 20/Jan/20
Commented by mind is power last updated on 21/Jan/20
2nd Way α^2 β^2 γ^2 δ^2 +α^2 β^2 +α^2 γ^2 +α^2 δ^2 +β^2 γ^2 +β^2 δ^2 +γ^2 δ^2 +α^2 β^2 γ^2 +α^2 β^2 δ^2 +β^2 γ^2 δ^2 +α^2 γ^2 δ^2 +1+α^2 +β^2 +δ^2 +γ^2 α^2 +β^2 +δ^2 +γ^2 =(α+β+δ+γ)^2 −2(αβ+βδ+βγ+αδ+αγ+δγ) =(−b)^2 −2c=b^2 −2c α^2 β^2 γ^2 δ^2 =(αβγδ)^2 =e^2 (α^2 β^2 +α^2 δ^2 +α^2 γ^2 +β^2 δ^2 +β^2 γ^2 +δ^2 γ^2 )=(αβ+αδ+αγ+βδ+βγ+δγ)^2 S=αβ+βγ+βδ+δγ+δα+γα −2α^2 βδ−2α^2 βγ−2α^2 δγ−2β^2 (αγ+γδ+αδ)−2γ^2 (αβ+αδ+δβ) −2δ^2 (αβ+αγ+γβ) =c^2 −2α^2 (S−α(β+γ+δ))−2β^2 (S−β(α+γ+δ))−2δ^2 (S−δ(α+β+γ)) =−2γ^2 (S−γ(α+β+δ)) =−2S(α^2 +β^2 +δ^2 +γ^2 )+2α^3 (β+γ+δ)+2β^3 (α+γ+δ)+2δ^2 (α+β+γ) +2γ^3 (α+β+δ) U=α+β+δ+γ =−2S(α^2 +β^2 +γ^2 +δ^2 )+2U(α^3 +β^3 +δ^3 +γ^3 )−2α^4 −2β^4 −2γ^4 −2δ^4 α^3 +β^3 +γ^3 +δ^3 use Newtoon Identity Too Bee continued not the Best Way to go for but Worcks
$$\mathrm{2}{nd}\:{Way} \\ $$$$\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} \delta^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \delta^{\mathrm{2}} +\beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\beta^{\mathrm{2}} \delta^{\mathrm{2}} +\gamma^{\mathrm{2}} \delta^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \delta^{\mathrm{2}} +\beta^{\mathrm{2}} \gamma^{\mathrm{2}} \delta^{\mathrm{2}} +\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} \delta^{\mathrm{2}} \\ $$$$+\mathrm{1}+\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\delta^{\mathrm{2}} +\gamma^{\mathrm{2}} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\delta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\left(\alpha+\beta+\delta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\beta\delta+\beta\gamma+\alpha\delta+\alpha\gamma+\delta\gamma\right) \\ $$$$=\left(−{b}\right)^{\mathrm{2}} −\mathrm{2}{c}={b}^{\mathrm{2}} −\mathrm{2}{c} \\ $$$$\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} \delta^{\mathrm{2}} =\left(\alpha\beta\gamma\delta\right)^{\mathrm{2}} ={e}^{\mathrm{2}} \\ $$$$\left(\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\alpha^{\mathrm{2}} \delta^{\mathrm{2}} +\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} +\beta^{\mathrm{2}} \delta^{\mathrm{2}} +\beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\delta^{\mathrm{2}} \gamma^{\mathrm{2}} \right)=\left(\alpha\beta+\alpha\delta+\alpha\gamma+\beta\delta+\beta\gamma+\delta\gamma\right)^{\mathrm{2}} \\ $$$${S}=\alpha\beta+\beta\gamma+\beta\delta+\delta\gamma+\delta\alpha+\gamma\alpha \\ $$$$−\mathrm{2}\alpha^{\mathrm{2}} \beta\delta−\mathrm{2}\alpha^{\mathrm{2}} \beta\gamma−\mathrm{2}\alpha^{\mathrm{2}} \delta\gamma−\mathrm{2}\beta^{\mathrm{2}} \left(\alpha\gamma+\gamma\delta+\alpha\delta\right)−\mathrm{2}\gamma^{\mathrm{2}} \left(\alpha\beta+\alpha\delta+\delta\beta\right) \\ $$$$−\mathrm{2}\delta^{\mathrm{2}} \left(\alpha\beta+\alpha\gamma+\gamma\beta\right) \\ $$$$={c}^{\mathrm{2}} −\mathrm{2}\alpha^{\mathrm{2}} \left({S}−\alpha\left(\beta+\gamma+\delta\right)\right)−\mathrm{2}\beta^{\mathrm{2}} \left({S}−\beta\left(\alpha+\gamma+\delta\right)\right)−\mathrm{2}\delta^{\mathrm{2}} \left({S}−\delta\left(\alpha+\beta+\gamma\right)\right) \\ $$$$=−\mathrm{2}\gamma^{\mathrm{2}} \left({S}−\gamma\left(\alpha+\beta+\delta\right)\right) \\ $$$$=−\mathrm{2}{S}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\delta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)+\mathrm{2}\alpha^{\mathrm{3}} \left(\beta+\gamma+\delta\right)+\mathrm{2}\beta^{\mathrm{3}} \left(\alpha+\gamma+\delta\right)+\mathrm{2}\delta^{\mathrm{2}} \left(\alpha+\beta+\gamma\right) \\ $$$$+\mathrm{2}\gamma^{\mathrm{3}} \left(\alpha+\beta+\delta\right) \\ $$$${U}=\alpha+\beta+\delta+\gamma \\ $$$$=−\mathrm{2}{S}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} \right)+\mathrm{2}{U}\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\delta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)−\mathrm{2}\alpha^{\mathrm{4}} −\mathrm{2}\beta^{\mathrm{4}} −\mathrm{2}\gamma^{\mathrm{4}} −\mathrm{2}\delta^{\mathrm{4}} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} +\delta^{\mathrm{3}} \\ $$$${use}\:{Newtoon}\:{Identity} \\ $$$${Too}\:{Bee}\:{continued}\:{not}\:{the}\:{Best}\:{Way}\:{to}\:{go}\:{for}\:{but}\:{Worcks} \\ $$
Answered by mind is power last updated on 20/Jan/20
(x^2 +1)=(x+i)(x−i) p(x)=x^4 +ax^3 +bx^2 +cx+d=(x−α)(x−β)(x−γ)(x−δ) (α^2 +1)(β^2 +1)(γ^2 +1)(δ^2 +1) =(α−i)(β−i)(γ−i)(δ−i)(α+i)(β+i)(γ+i)(δ+i) p(−i)=(−i−α)(−i−β)(−i−γ)(−i−δ) =(i+α)(i+β)(i+γ)(i+δ) p(i)=(i−α)(i−γ)(i−δ)(i−β)=(α−i)(β−i)(δ−i)(γ−i) ⇔(α^2 +1)(β^2 +1)(γ^2 +1)(δ^2 +1)=p(i).p(−i) =(1−ia−b+ci+d)(1+ia−b−ci+d) =(1−b+d+i(d−a))(1−b+d−i(d−a)) if b,c,d,e are reel =(1−b+d)^2 +(d−a)^2 other case=(1−b+d+i(d−a))(1−b+d−i(d−a))
$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)=\left(\mathrm{x}+\mathrm{i}\right)\left(\mathrm{x}−\mathrm{i}\right) \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{ax}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\left(\mathrm{x}−\alpha\right)\left(\mathrm{x}−\beta\right)\left(\mathrm{x}−\gamma\right)\left(\mathrm{x}−\delta\right) \\ $$$$\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)\left(\beta^{\mathrm{2}} +\mathrm{1}\right)\left(\gamma^{\mathrm{2}} +\mathrm{1}\right)\left(\delta^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\left(\alpha−\mathrm{i}\right)\left(\beta−\mathrm{i}\right)\left(\gamma−\mathrm{i}\right)\left(\delta−\mathrm{i}\right)\left(\alpha+\mathrm{i}\right)\left(\beta+\mathrm{i}\right)\left(\gamma+\mathrm{i}\right)\left(\delta+\mathrm{i}\right) \\ $$$$\mathrm{p}\left(−\mathrm{i}\right)=\left(−\mathrm{i}−\alpha\right)\left(−\mathrm{i}−\beta\right)\left(−\mathrm{i}−\gamma\right)\left(−\mathrm{i}−\delta\right) \\ $$$$=\left(\mathrm{i}+\alpha\right)\left(\mathrm{i}+\beta\right)\left(\mathrm{i}+\gamma\right)\left(\mathrm{i}+\delta\right) \\ $$$$\mathrm{p}\left(\mathrm{i}\right)=\left(\mathrm{i}−\alpha\right)\left(\mathrm{i}−\gamma\right)\left(\mathrm{i}−\delta\right)\left(\mathrm{i}−\beta\right)=\left(\alpha−\mathrm{i}\right)\left(\beta−\mathrm{i}\right)\left(\delta−\mathrm{i}\right)\left(\gamma−\mathrm{i}\right) \\ $$$$\Leftrightarrow\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)\left(\beta^{\mathrm{2}} +\mathrm{1}\right)\left(\gamma^{\mathrm{2}} +\mathrm{1}\right)\left(\delta^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{p}\left(\mathrm{i}\right).\mathrm{p}\left(−\mathrm{i}\right) \\ $$$$=\left(\mathrm{1}−\mathrm{ia}−\mathrm{b}+\mathrm{ci}+\mathrm{d}\right)\left(\mathrm{1}+\mathrm{ia}−\mathrm{b}−\mathrm{ci}+\mathrm{d}\right) \\ $$$$=\left(\mathrm{1}−\mathrm{b}+\mathrm{d}+\mathrm{i}\left(\mathrm{d}−\mathrm{a}\right)\right)\left(\mathrm{1}−\mathrm{b}+\mathrm{d}−\mathrm{i}\left(\mathrm{d}−\mathrm{a}\right)\right) \\ $$$$\mathrm{if}\:\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e}\:\mathrm{are}\:\mathrm{reel} \\ $$$$=\left(\mathrm{1}−\mathrm{b}+\mathrm{d}\right)^{\mathrm{2}} +\left(\mathrm{d}−\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{other}\:\mathrm{case}=\left(\mathrm{1}−\mathrm{b}+\mathrm{d}+\mathrm{i}\left(\mathrm{d}−\mathrm{a}\right)\right)\left(\mathrm{1}−\mathrm{b}+\mathrm{d}−\mathrm{i}\left(\mathrm{d}−\mathrm{a}\right)\right) \\ $$
Commented by M±th+et£s last updated on 20/Jan/20
nice job. though there is a small error in the fourth line from below when you want to factorize by i , its: i(c−a) and −i(c−a).
$${nice}\:{job}.\:{though}\:{there}\:{is}\:{a}\:{small}\:{error} \\ $$$${in}\:{the}\:{fourth}\:{line}\:{from}\:{below}\:{when} \\ $$$${you}\:{want}\:{to}\:{factorize}\:{by}\:{i}\:,\:{its}:\:\mathrm{i}\left(\mathrm{c}−\mathrm{a}\right)\:{and}\:−\mathrm{i}\left(\mathrm{c}−\mathrm{a}\right). \\ $$
Commented by mind is power last updated on 20/Jan/20
yeah
$$\mathrm{yeah} \\ $$
Commented by jagoll last updated on 21/Jan/20
mr Mind is power if we work using Vieta′s rule, given the same result?
$$\mathrm{mr}\:\mathrm{Mind}\:\mathrm{is}\:\mathrm{power} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{work}\:\mathrm{using}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{rule}, \\ $$$$\mathrm{given}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result}? \\ $$
Commented by mind is power last updated on 21/Jan/20
yeah
$$\mathrm{yeah}\:\: \\ $$

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