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Question Number 195628 by mr W last updated on 06/Aug/23
an unsolved old question #190875 a, b, c are real roots of the equation x^3 −7x^2 +4x+1=0. find (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=?
$$\underline{{an}\:{unsolved}\:{old}\:{question}\:#\mathrm{190875}} \\ $$$${a},\:{b},\:{c}\:{are}\:{real}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{0}. \\ $$$${find}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=? \\ $$
Commented by Frix last updated on 06/Aug/23
I don′t think we can give the exact value
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{give}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{value} \\ $$
Commented by mr W last updated on 06/Aug/23
but can we express the result in an exact form? just like sin (π/(13)), it is an exact form, even when we don′t know its exact value.
$${but}\:{can}\:{we}\:{express}\:{the}\:{result}\:{in}\:{an} \\ $$$${exact}\:{form}?\:{just}\:{like}\:\mathrm{sin}\:\frac{\pi}{\mathrm{13}},\:{it}\:{is} \\ $$$${an}\:{exact}\:{form},\:{even}\:{when}\:{we}\:{don}'{t} \\ $$$${know}\:{its}\:{exact}\:{value}. \\ $$
Commented by Frix last updated on 06/Aug/23
In this case (as you certainly know) we get 3 trigonometric solutions. I don′t think we can simplify Σ (1/( (x_j )^(1/3) )) We′d need to find a 3^(rd) degree factor of (1/y^9 )−(7/y^6 )+(4/y^3 )+1=0 ⇔ y^9 +4y^6 −7y^3 +1=0 with 3 real roots and I don′t think that′s possible.
$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\left(\mathrm{as}\:\mathrm{you}\:\mathrm{certainly}\:\mathrm{know}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{3}\:\mathrm{trigonometric}\:\mathrm{solutions}.\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{simplify}\:\Sigma\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{x}_{{j}} }} \\ $$$$\mathrm{We}'\mathrm{d}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}\:\mathrm{factor}\:\mathrm{of} \\ $$$$\frac{\mathrm{1}}{{y}^{\mathrm{9}} }−\frac{\mathrm{7}}{{y}^{\mathrm{6}} }+\frac{\mathrm{4}}{{y}^{\mathrm{3}} }+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{y}^{\mathrm{9}} +\mathrm{4}{y}^{\mathrm{6}} −\mathrm{7}{y}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{with}\:\mathrm{3}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{and}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}'\mathrm{s} \\ $$$$\mathrm{possible}. \\ $$
Answered by witcher3 last updated on 06/Aug/23
(1/( (a)^(1/3) ))..?
$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}..? \\ $$
Commented by mr W last updated on 06/Aug/23
yes!
$${yes}! \\ $$
Answered by mr W last updated on 06/Aug/23
a, b, c are roots of x^3 −7x^2 +4x+1=0. ((1/x))^3 +4((1/x))^2 −7((1/x))+1=0 so (1/a), (1/b), (1/c) are roots of z^3 +4z^2 −7z+1=0. let p=(1/a),q=(1/b),r=(1/c) p+q+r=−4 pq+qr+rp=−7 pqr=−1 say s=(1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=(p)^(1/3) +(q)^(1/3) +(r)^(1/3) s^3 =p+q+r−3((pqr))^(1/3) +3s(((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) ) ⇒s^3 =−1+3st ...(i) with t=((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) t^3 =pq+qr+rp−3(((pqr)^2 ))^(1/3) +3t((pqr))^(1/3) ((p)^(1/3) +(q)^(1/3) +(r)^(1/3) ) ⇒t^3 =−10−3ts ...(ii) (i)+(ii): s^3 +t^3 =−11 ⇒t^3 =−11−s^3 (i): s^3 +1=3st (s^3 +1)^3 =27s^3 t^3 (s^3 +1)^3 =27s^3 (−11−s^3 ) let u=s^3 (u+1)^3 =27u(−11−u) u^3 +30u^2 +300u+1=0 let u=v−10 (v−10)^3 +30(v−10)^2 +300(v−10)+1=0 v^3 −999=0 ⇒v=((999))^(1/3) ⇒s^3 =u=v−10=((999))^(1/3) −10 ⇒s=((((999))^(1/3) −10))^(1/3) i.e. (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=−((10−((999))^(1/3) ))^(1/3) ✓
$${a},\:{b},\:{c}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{0}. \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\mathrm{4}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{7}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}=\mathrm{0} \\ $$$${so}\:\frac{\mathrm{1}}{{a}},\:\frac{\mathrm{1}}{{b}},\:\frac{\mathrm{1}}{{c}}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{3}} +\mathrm{4}{z}^{\mathrm{2}} −\mathrm{7}{z}+\mathrm{1}=\mathrm{0}. \\ $$$${let}\:{p}=\frac{\mathrm{1}}{{a}},{q}=\frac{\mathrm{1}}{{b}},{r}=\frac{\mathrm{1}}{{c}} \\ $$$${p}+{q}+{r}=−\mathrm{4} \\ $$$${pq}+{qr}+{rp}=−\mathrm{7} \\ $$$${pqr}=−\mathrm{1} \\ $$$${say}\:{s}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{q}}+\sqrt[{\mathrm{3}}]{{r}} \\ $$$${s}^{\mathrm{3}} ={p}+{q}+{r}−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}+\mathrm{3}{s}\left(\sqrt[{\mathrm{3}}]{{pq}}+\sqrt[{\mathrm{3}}]{{qr}}+\sqrt[{\mathrm{3}}]{{rp}}\right) \\ $$$$\Rightarrow{s}^{\mathrm{3}} =−\mathrm{1}+\mathrm{3}{st}\:\:\:…\left({i}\right) \\ $$$${with}\:{t}=\sqrt[{\mathrm{3}}]{{pq}}+\sqrt[{\mathrm{3}}]{{qr}}+\sqrt[{\mathrm{3}}]{{rp}} \\ $$$${t}^{\mathrm{3}} ={pq}+{qr}+{rp}−\mathrm{3}\sqrt[{\mathrm{3}}]{\left({pqr}\right)^{\mathrm{2}} }+\mathrm{3}{t}\sqrt[{\mathrm{3}}]{{pqr}}\left(\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{q}}+\sqrt[{\mathrm{3}}]{{r}}\right) \\ $$$$\Rightarrow{t}^{\mathrm{3}} =−\mathrm{10}−\mathrm{3}{ts}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${s}^{\mathrm{3}} +{t}^{\mathrm{3}} =−\mathrm{11}\:\Rightarrow{t}^{\mathrm{3}} =−\mathrm{11}−{s}^{\mathrm{3}} \\ $$$$\left({i}\right):\: \\ $$$${s}^{\mathrm{3}} +\mathrm{1}=\mathrm{3}{st} \\ $$$$\left({s}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27}{s}^{\mathrm{3}} {t}^{\mathrm{3}} \\ $$$$\left({s}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27}{s}^{\mathrm{3}} \left(−\mathrm{11}−{s}^{\mathrm{3}} \right) \\ $$$${let}\:{u}={s}^{\mathrm{3}} \\ $$$$\left({u}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27}{u}\left(−\mathrm{11}−{u}\right) \\ $$$${u}^{\mathrm{3}} +\mathrm{30}{u}^{\mathrm{2}} +\mathrm{300}{u}+\mathrm{1}=\mathrm{0} \\ $$$${let}\:{u}={v}−\mathrm{10} \\ $$$$\left({v}−\mathrm{10}\right)^{\mathrm{3}} +\mathrm{30}\left({v}−\mathrm{10}\right)^{\mathrm{2}} +\mathrm{300}\left({v}−\mathrm{10}\right)+\mathrm{1}=\mathrm{0} \\ $$$${v}^{\mathrm{3}} −\mathrm{999}=\mathrm{0} \\ $$$$\Rightarrow{v}=\sqrt[{\mathrm{3}}]{\mathrm{999}} \\ $$$$\Rightarrow{s}^{\mathrm{3}} ={u}={v}−\mathrm{10}=\sqrt[{\mathrm{3}}]{\mathrm{999}}−\mathrm{10} \\ $$$$\Rightarrow{s}=\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{999}}−\mathrm{10}} \\ $$$${i}.{e}.\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=−\sqrt[{\mathrm{3}}]{\mathrm{10}−\sqrt[{\mathrm{3}}]{\mathrm{999}}}\:\checkmark \\ $$
Commented by Frix last updated on 06/Aug/23
Nice!
$$\mathrm{Nice}! \\ $$

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