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Question Number 210310 by Spillover last updated on 05/Aug/24
Let a be the unique real zero of x^3 +x+1. find the simplest possible way to write ((18)/((a^2 +a+1)^2 )) as polynomial expression in a with ratio coefficients
$${Let}\:{a}\:{be}\:{the}\:{unique}\:{real}\:{zero}\:{of}\:{x}^{\mathrm{3}} +{x}+\mathrm{1}. \\ $$$${find}\:{the}\:{simplest}\:{possible}\:{way}\:{to}\:{write}\: \\ $$$$\frac{\mathrm{18}}{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{as}\:{polynomial}\:{expression}\:{in}\:\:{a} \\ $$$${with}\:{ratio}\:{coefficients} \\ $$
Commented by Frix last updated on 07/Aug/24
Do you know the answer?
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}? \\ $$
Commented by AlagaIbile last updated on 07/Aug/24
6(a^2 + 1)^3
$$\:\:\mathrm{6}\left({a}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{3}} \\ $$
Commented by Frix last updated on 08/Aug/24
((18)/((a^2 +a+1)^2 ))≈29.34 6(a^2 +1)^3 ≈18.89
$$\frac{\mathrm{18}}{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)^{\mathrm{2}} }\approx\mathrm{29}.\mathrm{34} \\ $$$$\mathrm{6}\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \approx\mathrm{18}.\mathrm{89} \\ $$
Commented by Spillover last updated on 08/Aug/24
14a^2 −10a+16
$$\mathrm{14}{a}^{\mathrm{2}} −\mathrm{10}{a}+\mathrm{16} \\ $$
Answered by Frix last updated on 08/Aug/24
We have a^3 +a+1=0∧a∈R ((18)/((a^2 +a+1)^2 ))=((18)/(a^4 +2a^3 +3a^2 +2a+1))= a^3 =−(a+1) ⇒ a^4 =−a(a+1) =((18)/(2a^3 +2a^2 +a+1))= a^3 =−(a+1) =((18)/(2a^2 −a−1))= a^2 =−((a+1)/a) =−((18a)/(a^2 +3a+2))=−((18a)/((a+1)(a+2)))= =((18)/(a+1))−((36)/(a+2)) We know a=u^(1/3) +v^(1/3) ω=−(1/2)+((√3)/2)i (p/(a+q))=(p/(q+u^(1/3) +v^(1/3) ))= =(p/(q+u^(1/3) +v^(1/3) ))×(((q+ωu^(1/3) +ω^2 v^(1/3) )(q+ω^2 u^(1/3) +ωv^(1/3) ))/((q+ωu^(1/3) +ω^2 v^(1/3) )(q+ω^2 u^(1/3) +ωv^(1/3) )))= =((p(q^2 −(u^(1/3) +v^(1/3) )q+u^(2/3) −u^(1/3) v^(1/3) +v^(2/3) ))/(q^3 +u+v−3u^(1/3) v^(1/3) q))= Cardano gives u, v: a=(−(1/2)+((√(93))/(18)))^(1/3) +(−(1/2)−((√(93))/(18)))^(1/3) u=−(1/2)+((√(93))/(18))∧v=−(1/2)−((√(93))/(18)) uv=−(1/(27)) ⇒ u+v=−1∧u^(1/3) v^(1/3) =−(1/3) =((p(q^2 −qa+u^(2/3) +2u^(1/3) v^(1/3) +v^(2/3) −3u^(1/3) v^(1/3) ))/(q^3 +q−1))= =((p(q^2 −qa+a^2 +1))/(q^3 +q−1)) ((18)/(a+1))−((36)/(a+2))=18(x^2 −x+2)−4(x^2 −2x+5)= =2(7x^2 −5x+8)
$$\mathrm{We}\:\mathrm{have}\:{a}^{\mathrm{3}} +{a}+\mathrm{1}=\mathrm{0}\wedge{a}\in\mathbb{R} \\ $$$$\frac{\mathrm{18}}{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{18}}{{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}= \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} =−\left({a}+\mathrm{1}\right)\:\Rightarrow\:{a}^{\mathrm{4}} =−{a}\left({a}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{18}}{\mathrm{2}{a}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{2}} +{a}+\mathrm{1}}= \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} =−\left({a}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{18}}{\mathrm{2}{a}^{\mathrm{2}} −{a}−\mathrm{1}}= \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} =−\frac{{a}+\mathrm{1}}{{a}} \\ $$$$=−\frac{\mathrm{18}{a}}{{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{2}}=−\frac{\mathrm{18}{a}}{\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)}= \\ $$$$=\frac{\mathrm{18}}{{a}+\mathrm{1}}−\frac{\mathrm{36}}{{a}+\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{know}\:{a}={u}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\frac{{p}}{{a}+{q}}=\frac{{p}}{{q}+{u}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{1}}{\mathrm{3}}} }= \\ $$$$=\frac{{p}}{{q}+{u}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{1}}{\mathrm{3}}} }×\frac{\left({q}+\omega{u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\omega^{\mathrm{2}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({q}+\omega^{\mathrm{2}} {u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\omega{v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}{\left({q}+\omega{u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\omega^{\mathrm{2}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({q}+\omega^{\mathrm{2}} {u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\omega{v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}= \\ $$$$=\frac{{p}\left({q}^{\mathrm{2}} −\left({u}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right){q}+{u}^{\frac{\mathrm{2}}{\mathrm{3}}} −{u}^{\frac{\mathrm{1}}{\mathrm{3}}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}{{q}^{\mathrm{3}} +{u}+{v}−\mathrm{3}{u}^{\frac{\mathrm{1}}{\mathrm{3}}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} {q}}= \\ $$$$\:\:\:\:\:\mathrm{Cardano}\:\mathrm{gives}\:{u},\:{v}: \\ $$$$\:\:\:\:\:{a}=\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:{u}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}\wedge{v}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{93}}}{\mathrm{18}} \\ $$$$\:\:\:\:\:{uv}=−\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow\:{u}+{v}=−\mathrm{1}\wedge{u}^{\frac{\mathrm{1}}{\mathrm{3}}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} =−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{{p}\left({q}^{\mathrm{2}} −{qa}+{u}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}{u}^{\frac{\mathrm{1}}{\mathrm{3}}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{3}{u}^{\frac{\mathrm{1}}{\mathrm{3}}} {v}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}{{q}^{\mathrm{3}} +{q}−\mathrm{1}}= \\ $$$$=\frac{{p}\left({q}^{\mathrm{2}} −{qa}+{a}^{\mathrm{2}} +\mathrm{1}\right)}{{q}^{\mathrm{3}} +{q}−\mathrm{1}} \\ $$$$ \\ $$$$\frac{\mathrm{18}}{{a}+\mathrm{1}}−\frac{\mathrm{36}}{{a}+\mathrm{2}}=\mathrm{18}\left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)−\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)= \\ $$$$=\mathrm{2}\left(\mathrm{7}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}\right) \\ $$
Answered by Spillover last updated on 08/Aug/24

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