Question Number 128150 by MJS_new last updated on 04/Jan/21
![question 128091, trying to solve completely in R a^3 =3(b^2 +c^2 )−25 b^3 =3(a^2 +c^2 )−25 c^3 =3(a^2 +b^2 )−25 abc=? (1) a=b=c a^3 −6a^3 +25=0 (a−5)(a^2 −a−5)=0 a=5∨a=(1/2)−((√(21))/2)∨a=(1/2)+((√(21))/2) (2) c=b≠a a^3 =6b^2 −25 ⇒ a=((6b^2 −25))^(1/3) b^3 =3(a^2 +b^2 )−25 b^3 −3b^2 +25=3(6b^2 −25)^(2/3) b^9 −9b^8 +27b^7 +48b^6 −450b^5 −297b^4 +1875b^3 +2475b^2 −1250=0 (b−5)(b^2 −b−5)(b^6 −3b^5 +9b^4 +77b^3 +87b^2 −50)=0 the 1^(st) and 2^(nd) brackets are included in (1) b^6 −3b^5 +9b^4 +77b^3 +87b^2 −50=0 b_1 ≈.605473087 b_2 ≈−2.08139378 a_1 ≈−2.83561704 a_2 ≈.997728331 (3) a≠b≠c let b=pa∧c=qa a^3 =3a^2 (p^2 +q^2 )−25 a^3 p^3 =3a^2 (q^2 +1)−25 a^3 q^3 =3a^2 (p^2 +1)−25 3a^2 (p^2 +q^2 )−25=((3a^2 (q^2 +1)−25)/p^3 ) 3a^2 (p^2 +q^2 )−25=((3a^2 (p^2 +1)−25)/q^3 ) ⇒ [p=1∧q=1 included in (1)∨(2)] a^2 =((25(p^2 +p+1))/(3(p^4 +p^3 +p^2 q^2 +p^2 +pq^2 +p+q^2 +1))) a^2 =((25(q^2 +q+1))/(3(p^2 q^2 +p^2 q+p^2 +q^4 +q^3 +q^2 +q+1))) equating and transforming ⇒ q^2 −(p^2 /(p+1))q−(p^2 /(p+1))=0 q=−(p/(p+1)) [q=p included in (1)∨(2)] ⇒ a^2 =((25(p+1)^2 (p^2 +p+1))/(3(p^6 +3p^5 +5p^4 +5p^3 +5p^2 +3p+1))) inserting above leads to (assuming a<b<c) a=−1−(√3)∧b=−1∧c=−1+(√3) all solutions interchangeable a⇆b⇆c](Q128150.png)
$$\mathrm{question}\:\mathrm{128091},\:\mathrm{trying}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{completely}\:\mathrm{in}\:\mathbb{R} \\ $$$${a}^{\mathrm{3}} =\mathrm{3}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${b}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${c}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${abc}=? \\ $$$$\left(\mathrm{1}\right)\:{a}={b}={c} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{3}} +\mathrm{25}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({a}−\mathrm{5}\right)\left({a}^{\mathrm{2}} −{a}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{a}=\mathrm{5}\vee{a}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{21}}}{\mathrm{2}}\vee{a}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{c}={b}\neq{a} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} =\mathrm{6}{b}^{\mathrm{2}} −\mathrm{25}\:\Rightarrow\:{a}=\sqrt[{\mathrm{3}}]{\mathrm{6}{b}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\:\:\:\:\:{b}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$$ \\ $$$$\:\:\:\:\:{b}^{\mathrm{3}} −\mathrm{3}{b}^{\mathrm{2}} +\mathrm{25}=\mathrm{3}\left(\mathrm{6}{b}^{\mathrm{2}} −\mathrm{25}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\:\:\:\:\:{b}^{\mathrm{9}} −\mathrm{9}{b}^{\mathrm{8}} +\mathrm{27}{b}^{\mathrm{7}} +\mathrm{48}{b}^{\mathrm{6}} −\mathrm{450}{b}^{\mathrm{5}} −\mathrm{297}{b}^{\mathrm{4}} +\mathrm{1875}{b}^{\mathrm{3}} +\mathrm{2475}{b}^{\mathrm{2}} −\mathrm{1250}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({b}−\mathrm{5}\right)\left({b}^{\mathrm{2}} −{b}−\mathrm{5}\right)\left({b}^{\mathrm{6}} −\mathrm{3}{b}^{\mathrm{5}} +\mathrm{9}{b}^{\mathrm{4}} +\mathrm{77}{b}^{\mathrm{3}} +\mathrm{87}{b}^{\mathrm{2}} −\mathrm{50}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{brackets}\:\mathrm{are}\:\mathrm{included}\:\mathrm{in}\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:{b}^{\mathrm{6}} −\mathrm{3}{b}^{\mathrm{5}} +\mathrm{9}{b}^{\mathrm{4}} +\mathrm{77}{b}^{\mathrm{3}} +\mathrm{87}{b}^{\mathrm{2}} −\mathrm{50}=\mathrm{0} \\ $$$$\:\:\:\:\:{b}_{\mathrm{1}} \approx.\mathrm{605473087}\:\:\:\:\:{b}_{\mathrm{2}} \approx−\mathrm{2}.\mathrm{08139378} \\ $$$$\:\:\:\:\:{a}_{\mathrm{1}} \approx−\mathrm{2}.\mathrm{83561704}\:\:\:\:\:{a}_{\mathrm{2}} \approx.\mathrm{997728331} \\ $$$$\left(\mathrm{3}\right)\:{a}\neq{b}\neq{c} \\ $$$$\:\:\:\:\:\mathrm{let}\:{b}={pa}\wedge{c}={qa} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} =\mathrm{3}{a}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} {p}^{\mathrm{3}} =\mathrm{3}{a}^{\mathrm{2}} \left({q}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{25} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} {q}^{\mathrm{3}} =\mathrm{3}{a}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{25} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{3}{a}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{25}=\frac{\mathrm{3}{a}^{\mathrm{2}} \left({q}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{25}}{{p}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\mathrm{3}{a}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{25}=\frac{\mathrm{3}{a}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{25}}{{q}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\Rightarrow\:\left[{p}=\mathrm{1}\wedge{q}=\mathrm{1}\:\mathrm{included}\:\mathrm{in}\:\left(\mathrm{1}\right)\vee\left(\mathrm{2}\right)\right] \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} =\frac{\mathrm{25}\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\mathrm{3}\left({p}^{\mathrm{4}} +{p}^{\mathrm{3}} +{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} +{pq}^{\mathrm{2}} +{p}+{q}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} =\frac{\mathrm{25}\left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)}{\mathrm{3}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {q}+{p}^{\mathrm{2}} +{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}^{\mathrm{2}} +{q}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\mathrm{equating}\:\mathrm{and}\:\mathrm{transforming}\:\Rightarrow \\ $$$$\:\:\:\:\:{q}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{{p}+\mathrm{1}}{q}−\frac{{p}^{\mathrm{2}} }{{p}+\mathrm{1}}=\mathrm{0} \\ $$$$\:\:\:\:\:{q}=−\frac{{p}}{{p}+\mathrm{1}}\:\:\:\:\:\left[{q}={p}\:\mathrm{included}\:\mathrm{in}\:\left(\mathrm{1}\right)\vee\left(\mathrm{2}\right)\right] \\ $$$$\:\:\:\:\:\Rightarrow\:{a}^{\mathrm{2}} =\frac{\mathrm{25}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\mathrm{3}\left({p}^{\mathrm{6}} +\mathrm{3}{p}^{\mathrm{5}} +\mathrm{5}{p}^{\mathrm{4}} +\mathrm{5}{p}^{\mathrm{3}} +\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\mathrm{inserting}\:\mathrm{above}\:\mathrm{leads}\:\mathrm{to}\:\left(\mathrm{assuming}\:{a}<{b}<{c}\right) \\ $$$$\:\:\:\:\:{a}=−\mathrm{1}−\sqrt{\mathrm{3}}\wedge{b}=−\mathrm{1}\wedge{c}=−\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{all}\:\mathrm{solutions}\:\mathrm{interchangeable}\:{a}\leftrightarrows{b}\leftrightarrows{c} \\ $$
Commented by MJS_new last updated on 04/Jan/21
$$\mathrm{didn}'\mathrm{t}\:\mathrm{try}\:\mathrm{in}\:\mathbb{C}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{very}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{do} \\ $$
Commented by Tawa11 last updated on 06/Nov/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by rydasss last updated on 05/Jan/21
$${thank}\:{you}\:{so}\:{much} \\ $$