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Question Number 143602 by meetbhavsar25 last updated on 16/Jun/21

Commented by TheHoneyCat last updated on 16/Jun/21

Are you asking for Π_(a∈C) g(a)^(val_f (a)) or z such that ∀(a_i )_(i∈[∣1,6∣]) Π_(i=1) ^6 g(a_i )=z ¿ (that is to say, is each a_i one sigle root of f or can they all be the same for instance)

$$\mathrm{Are}\:\mathrm{you}\:\mathrm{asking}\:\mathrm{for}\:\underset{{a}\in\mathbb{C}} {\prod}{g}\left({a}\right)^{\mathrm{val}_{{f}} \left({a}\right)} \\ $$$$ \\ $$$$\mathrm{or}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\forall\left({a}_{{i}} \right)_{{i}\in\left[\mid\mathrm{1},\mathrm{6}\mid\right]} \:\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{g}\left({a}_{{i}} \right)={z}\: \\ $$$$ \\ $$$$¿ \\ $$$$ \\ $$$$\left({that}\:{is}\:{to}\:{say},\:{is}\:{each}\:{a}_{{i}} \:{one}\:{sigle}\:{root}\:{of}\:{f}\:{or}\:{can}\:{they}\:{all}\:{be}\:{the}\:{same}\:{for}\:{instance}\right) \\ $$

Answered by mr W last updated on 16/Jun/21

f(x)=(x^3 +2)(x^3 −4)=0 ⇒α_1 ^3 ,α_2 ^3 ,α_3 ^3 =−2 ⇒α_(1,2,) =−(2)^(1/3) ,−(2)^(1/3) ω,−(2)^(1/3) ω^2 ⇒α_4 ^3 ,α_5 ^3 ,α_6 ^3 =4 ⇒α_(4,5,6) =(4)^(1/3) ,(4)^(1/3) ω,(4)^(1/3) ω^2 g(x)=(x+1+i(√3))(x+1−i(√3)) g(x)=(x−2ω)(x−2ω^2 ) Π_(k=1) ^6 g(α_k )=Π_(k=1) ^6 (α_k −2ω)(α_k −2ω^2 ) =(−(2)^(1/3) −2ω)(−(2)^(1/3) −2ω^2 ) ×(−(2)^(1/3) ω−2ω)(−(2)^(1/3) ω−2ω^2 ) ×(−(2)^(1/3) ω^2 −2ω)(−(2)^(1/3) ω^2 −2ω^2 ) ×((4)^(1/3) −2ω)((4)^(1/3) −2ω^2 ) ×((4)^(1/3) ω−2ω)((4)^(1/3) ω−2ω^2 ) ×((4)^(1/3) ω^2 −2ω)((4)^(1/3) ω^2 −2ω^2 ) =((2)^(1/3) +2ω)((2)^(1/3) ω+2)((2)^(1/3) +2)^2 ((2)^(1/3) +2ω)((2)^(1/3) ω+2) ×((4)^(1/3) −2ω)((4)^(1/3) ω−2)((4)^(1/3) −2)^2 ((4)^(1/3) −2ω)((4)^(1/3) ω−2)ω^2 =((2)^(1/3) +2ω)^2 ((2)^(1/3) ω+2)^2 ((2)^(1/3) +2)^2 ×((4)^(1/3) −2ω)^2 ((4)^(1/3) ω−2)^2 ((4)^(1/3) −2)^2 ω^2 =1600

$${f}\left({x}\right)=\left({x}^{\mathrm{3}} +\mathrm{2}\right)\left({x}^{\mathrm{3}} −\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\alpha_{\mathrm{1}} ^{\mathrm{3}} ,\alpha_{\mathrm{2}} ^{\mathrm{3}} ,\alpha_{\mathrm{3}} ^{\mathrm{3}} =−\mathrm{2} \\ $$$$\Rightarrow\alpha_{\mathrm{1},\mathrm{2},} =−\sqrt[{\mathrm{3}}]{\mathrm{2}},−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega,−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega^{\mathrm{2}} \\ $$$$\Rightarrow\alpha_{\mathrm{4}} ^{\mathrm{3}} ,\alpha_{\mathrm{5}} ^{\mathrm{3}} ,\alpha_{\mathrm{6}} ^{\mathrm{3}} =\mathrm{4} \\ $$$$\Rightarrow\alpha_{\mathrm{4},\mathrm{5},\mathrm{6}} =\sqrt[{\mathrm{3}}]{\mathrm{4}},\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega,\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega^{\mathrm{2}} \\ $$$${g}\left({x}\right)=\left({x}+\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)\left({x}+\mathrm{1}−{i}\sqrt{\mathrm{3}}\right) \\ $$$${g}\left({x}\right)=\left({x}−\mathrm{2}\omega\right)\left({x}−\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{g}\left(\alpha_{{k}} \right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\left(\alpha_{{k}} −\mathrm{2}\omega\right)\left(\alpha_{{k}} −\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$=\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{2}\omega\right)\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$×\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega−\mathrm{2}\omega\right)\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega−\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$×\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega^{\mathrm{2}} −\mathrm{2}\omega\right)\left(−\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega^{\mathrm{2}} −\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$×\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$×\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega−\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega−\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$×\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega^{\mathrm{2}} −\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega^{\mathrm{2}} −\mathrm{2}\omega^{\mathrm{2}} \right) \\ $$$$=\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega+\mathrm{2}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega+\mathrm{2}\right) \\ $$$$×\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega−\mathrm{2}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\omega\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega−\mathrm{2}\right)\omega^{\mathrm{2}} \\ $$$$=\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{2}\omega\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega+\mathrm{2}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$×\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\omega\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\omega−\mathrm{2}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\right)^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$=\mathrm{1600} \\ $$

Answered by ajfour last updated on 16/Jun/21

x^6 −2x^3 −8=0 and x^2 =g−2x−4 (g−2x−4)^3 −2x(g−2x−4)=8 (g−4)^3 −8x(g−2x−4) −6(g−4)^2 x+12(g−4)(g−2x−4) −2(g−4)x+4(g−2x−4)−8=0 ⇒ (g−4)^3 +16(g−2x−4) −8x(g−4)−6(g−4)^2 x +12(g−4)^2 −24(g−4)x −2(g−4)x+4(g−4)−8x−8 =0 ⇒ x=(((g−4)^3 +12(g−4)^2 +20(g−4)−8)/(6(g−4)^2 +34(g−4)+40)) ⇒ but (x+1)^2 =g−3 ⇒ {(((g−4)^3 +18(g−4)^2 +54(g−4)+32)/(6(g−4)^2 +34(g−4)+40))}^2 =g−3 Πg_i ={(−4)^3 +18(−4)^2 +54(−4)+32}^2 −(−3){6(−4)^2 +34(−4)+40}^2 =16(−16+72−54+8)^2 +48(24−34+10)^2 = 1600 ★ yes sir!

$${x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{8}=\mathrm{0} \\ $$$${and}\:\:\:{x}^{\mathrm{2}} ={g}−\mathrm{2}{x}−\mathrm{4} \\ $$$$\left({g}−\mathrm{2}{x}−\mathrm{4}\right)^{\mathrm{3}} −\mathrm{2}{x}\left({g}−\mathrm{2}{x}−\mathrm{4}\right)=\mathrm{8} \\ $$$$\left({g}−\mathrm{4}\right)^{\mathrm{3}} −\mathrm{8}{x}\left({g}−\mathrm{2}{x}−\mathrm{4}\right) \\ $$$$−\mathrm{6}\left({g}−\mathrm{4}\right)^{\mathrm{2}} {x}+\mathrm{12}\left({g}−\mathrm{4}\right)\left({g}−\mathrm{2}{x}−\mathrm{4}\right) \\ $$$$−\mathrm{2}\left({g}−\mathrm{4}\right){x}+\mathrm{4}\left({g}−\mathrm{2}{x}−\mathrm{4}\right)−\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow\: \\ $$$$\left({g}−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{16}\left({g}−\mathrm{2}{x}−\mathrm{4}\right) \\ $$$$−\mathrm{8}{x}\left({g}−\mathrm{4}\right)−\mathrm{6}\left({g}−\mathrm{4}\right)^{\mathrm{2}} {x} \\ $$$$+\mathrm{12}\left({g}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{24}\left({g}−\mathrm{4}\right){x} \\ $$$$−\mathrm{2}\left({g}−\mathrm{4}\right){x}+\mathrm{4}\left({g}−\mathrm{4}\right)−\mathrm{8}{x}−\mathrm{8} \\ $$$$=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{\left({g}−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{12}\left({g}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{20}\left({g}−\mathrm{4}\right)−\mathrm{8}}{\mathrm{6}\left({g}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{34}\left({g}−\mathrm{4}\right)+\mathrm{40}} \\ $$$$\Rightarrow\:{but}\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} ={g}−\mathrm{3} \\ $$$$\Rightarrow\:\left\{\frac{\left({g}−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{18}\left({g}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{54}\left({g}−\mathrm{4}\right)+\mathrm{32}}{\mathrm{6}\left({g}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{34}\left({g}−\mathrm{4}\right)+\mathrm{40}}\right\}^{\mathrm{2}} \\ $$$$\:\:={g}−\mathrm{3} \\ $$$$\Pi{g}_{{i}} =\left\{\left(−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{18}\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{54}\left(−\mathrm{4}\right)+\mathrm{32}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:−\left(−\mathrm{3}\right)\left\{\mathrm{6}\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{34}\left(−\mathrm{4}\right)+\mathrm{40}\right\}^{\mathrm{2}} \\ $$$$=\mathrm{16}\left(−\mathrm{16}+\mathrm{72}−\mathrm{54}+\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:+\mathrm{48}\left(\mathrm{24}−\mathrm{34}+\mathrm{10}\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{1600}\:\:\bigstar \\ $$$${yes}\:{sir}! \\ $$$$ \\ $$

Commented by mr W last updated on 16/Jun/21

i got 1600. but my method is not as smart.

$${i}\:{got}\:\mathrm{1600}.\:{but}\:{my}\:{method}\:{is}\:{not}\:{as} \\ $$$${smart}. \\ $$

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