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


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Question Number 128150 by MJS_new last updated on 04/Jan/21

$question 128091, trying to solve completely in R$ $a^{3} = 3 (b^{2} + c^{2}) - 25$ $b^{3} = 3 (a^{2} + c^{2}) - 25$ $c^{3} = 3 (a^{2} + b^{2}) - 25$ $a b c = ?$ $(1) a = b = c$ $a^{3} - 6 a^{3} + 25 = 0$ $(a - 5) (a^{2} - a - 5) = 0$ $a = 5 \lor a = \frac{1}{2} - \frac{\sqrt{21}}{2} \lor a = \frac{1}{2} + \frac{\sqrt{21}}{2}$ $(2) c = b \neq a$ $a^{3} = 6 b^{2} - 25 \Rightarrow a = \sqrt[3]{6 b^{2} - 25}$ $b^{3} = 3 (a^{2} + b^{2}) - 25$ $b^{3} - 3 b^{2} + 25 = 3 {(6 b^{2} - 25)}^{2 / 3}$ $b^{9} - 9 b^{8} + 27 b^{7} + 48 b^{6} - 450 b^{5} - 297 b^{4} + 1875 b^{3} + 2475 b^{2} - 1250 = 0$ $(b - 5) (b^{2} - b - 5) (b^{6} - 3 b^{5} + 9 b^{4} + 77 b^{3} + 87 b^{2} - 50) = 0$ $the 1^{st} and 2^{nd} brackets are included in (1)$ $b^{6} - 3 b^{5} + 9 b^{4} + 77 b^{3} + 87 b^{2} - 50 = 0$ $b_{1} \approx . 605473087 b_{2} \approx - 2.08139378$ $a_{1} \approx - 2.83561704 a_{2} \approx . 997728331$ $(3) a \neq b \neq c$ $let b = p a \land c = q a$ $a^{3} = 3 a^{2} (p^{2} + q^{2}) - 25$ $a^{3} p^{3} = 3 a^{2} (q^{2} + 1) - 25$ $a^{3} q^{3} = 3 a^{2} (p^{2} + 1) - 25$ $3 a^{2} (p^{2} + q^{2}) - 25 = \frac{3 a^{2} (q^{2} + 1) - 25}{p^{3}}$ $3 a^{2} (p^{2} + q^{2}) - 25 = \frac{3 a^{2} (p^{2} + 1) - 25}{q^{3}}$ $\Rightarrow [p = 1 \land q = 1 included in (1) \lor (2)]$ $a^{2} = \frac{25 (p^{2} + p + 1)}{3 (p^{4} + p^{3} + p^{2} q^{2} + p^{2} + {p q}^{2} + p + q^{2} + 1)}$ $a^{2} = \frac{25 (q^{2} + q + 1)}{3 (p^{2} q^{2} + p^{2} q + p^{2} + q^{4} + q^{3} + q^{2} + q + 1)}$ $equating and transforming \Rightarrow$ $q^{2} - \frac{p^{2}}{p + 1} q - \frac{p^{2}}{p + 1} = 0$ $q = - \frac{p}{p + 1} [q = p included in (1) \lor (2)]$ $\Rightarrow a^{2} = \frac{25 {(p + 1)}^{2} (p^{2} + p + 1)}{3 (p^{6} + 3 p^{5} + 5 p^{4} + 5 p^{3} + 5 p^{2} + 3 p + 1)}$ $inserting above leads to (assuming a < b < c)$ $a = - 1 - \sqrt{3} \land b = - 1 \land c = - 1 + \sqrt{3}$ $all solutions interchangeable a ⇆ b ⇆ c$
Commented by MJS_new last updated on 04/Jan/21

${didn}^{'} t try in C because {it}^{'} s very hard to do$
Commented by Tawa11 last updated on 06/Nov/21

$Great sir$
	Answered by rydasss last updated on 05/Jan/21

	$t h a n k y o u s o m u c h$
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