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Question Number 70040 by Shamim last updated on 30/Sep/19
If, a^2 b^2 c^2 ((1/a^3 )+(1/b^3 )+(1/c^3 ))=a^3 +b^3 +c^3 than prove that a,b,c Successive Proportional.
If,a2b2c2(1a3+1b3+1c3)=a3+b3+c3thanprovethata,b,cSuccessiveProportional.
Commented by Prithwish sen last updated on 01/Oct/19
(((b^2 c^2 )/a)−a^3 )+(((a^2 c^2 )/b)−b^3 )+(((a^2 b^2 )/c) − c^3 )=0 ((b^2 c^2 −a^4 )/a) +((a^2 b^2 −c^4 )/c) +(((a^2 c^2 −b^4 )/c))=0 ((b^2 c^3 −a^4 c + a^3 b^2 −ac^4 )/(ac)) +((a^2 c^2 − b^4 )/c) = 0 (((b^2 −ac)(a^3 + c^3 ))/(ac)) − (((b^2 −ac)(b^2 +ac))/c) = 0 (b^2 −ac)(((a^3 +c^3 )/(ac)) − ((b^2 +ac)/c) )=0 considering only b^2 −ac = 0 ⇒ a,b,c are in proportion. proved
(b2c2a−a3)+(a2c2b−b3)+(a2b2c−c3)=0b2c2−a4a+a2b2−c4c+(a2c2−b4c)=0b2c3−a4c+a3b2−ac4ac+a2c2−b4c=0(b2−ac)(a3+c3)ac−(b2−ac)(b2+ac)c=0(b2−ac)(a3+c3ac−b2+acc)=0consideringonlyb2−ac=0⇒a,b,careinproportion.proved
Commented by Shamim last updated on 01/Oct/19
tnks
tnks
Commented by MJS last updated on 01/Oct/19
of course this is right I saw that the given equation can be transformed to a^4 bc+ab^4 c+abc^4 −a^3 b^3 −a^3 c^3 −b^3 c^3 =0 ⇔ ⇔ (a^2 −bc)(b^2 −ac)(c^2 −ab)=0 but I thought it′s not easy to factorize on the other hand by taking the result and simply putting b=pa, c=p^2 a or just c=(b^2 /a) we easily find the given equation is true but I thought showing that with b=pa and c=qa ⇒ q=p^2 is more convincing
ofcoursethisisrightIsawthatthegivenequationcanbetransformedtoa4bc+ab4c+abc4−a3b3−a3c3−b3c3=0⇔⇔(a2−bc)(b2−ac)(c2−ab)=0butIthoughtit′snoteasytofactorizeontheotherhandbytakingtheresultandsimplyputtingb=pa,c=p2aorjustc=b2aweeasilyfindthegivenequationistruebutIthoughtshowingthatwithb=paandc=qa⇒q=p2ismoreconvincing
Answered by MJS last updated on 30/Sep/19
for a, b, c ∈R we can find p, q ∈R ⇒ b=pa∧c=qa a^2 b^2 c^2 ((1/a^3 )+(1/b^3 )+(1/c^3 ))=a^3 +b^3 +c^3 a^3 b^3 +a^3 c^3 +b^3 c^3 =a^4 bc+ab^4 c+abc^4 a^4 bc−a^3 (b^3 −c^3 )+a(b^4 c+bc^4 )−b^3 c^3 =0 b=pa∧c=qa, let a≤b≤c ⇒ 1≤p≤q a^6 (−p^3 q^3 +p^4 q+pq^4 −p^3 −q^3 +pq)=0 a≠0 −p^3 q^3 +p^4 q+pq^4 −p^3 −q^3 +pq=0 (p^2 −q)(p−q^2 )(pq−1)=0 ⇒ q=p^2 [∨q=±(√p)∨q=(1/p)] ⇒ ⟨a, b, c⟩=⟨a, pa, p^2 a⟩
fora,b,c∈Rwecanfindp,q∈R⇒b=pa∧c=qaa2b2c2(1a3+1b3+1c3)=a3+b3+c3a3b3+a3c3+b3c3=a4bc+ab4c+abc4a4bc−a3(b3−c3)+a(b4c+bc4)−b3c3=0b=pa∧c=qa,leta⩽b⩽c⇒1⩽p⩽qa6(−p3q3+p4q+pq4−p3−q3+pq)=0a≠0−p3q3+p4q+pq4−p3−q3+pq=0(p2−q)(p−q2)(pq−1)=0⇒q=p2[∨q=±p∨q=1p]⇒⟨a,b,c⟩=⟨a,pa,p2a⟩
Commented by Shamim last updated on 30/Sep/19
It′s very hard. others rule???
It′sveryhard.othersrule???
Commented by MJS last updated on 30/Sep/19
it′s not that hard. write −p^3 q^3 +p^4 q+pq^4 −p^3 −q^3 +pq=0 as pq^4 −(p^3 +1)q^3 +p(p^3 +1)q−p^3 =0 divide by p q^4 −((p^3 +1)/p)q^3 +(p^3 +1)q−p^2 =0 we know if there′s an easy solution q_1 ⇒ q_1 ∣p^2 ⇒ we have to try ±1, ±p, ±p^2 ⇒ q_1 =p^2 and this is enough because we set a≤b≤c ⇒ 1≤p≤q ⇒ q≠±(√p)∧q≠(1/p) we then have (q−p^2 )(q^3 −(1/p)q^2 −pq+1)=0 transform to (q−p^2 )(q^2 (q−(1/p))−p(q−(1/p)))=0 ⇒ q_2 =(1/p) (q−p^2 )(q−(1/p))(q^2 −p)=0 ⇒ q_(3, 4) =±(√p)
it′snotthathard.write−p3q3+p4q+pq4−p3−q3+pq=0aspq4−(p3+1)q3+p(p3+1)q−p3=0dividebypq4−p3+1pq3+(p3+1)q−p2=0weknowifthere′saneasysolutionq1⇒q1∣p2⇒wehavetotry±1,±p,±p2⇒q1=p2andthisisenoughbecauseweseta⩽b⩽c⇒1⩽p⩽q⇒q≠±p∧q≠1pwethenhave(q−p2)(q3−1pq2−pq+1)=0transformto(q−p2)(q2(q−1p)−p(q−1p))=0⇒q2=1p(q−p2)(q−1p)(q2−p)=0⇒q3,4=±p

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