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If-a-b-and-c-are-the-roots-of-x-3-21x-35-0-what-is-the-value-of-a-b-c-a-b-c-

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Question Number 129336 by bramlexs22 last updated on 15/Jan/21
If a,b and c are the roots of x^3 −21x−35 = 0 what is the value of (a−b)(c−a)(b−c) ?
Ifa,bandcaretherootsofx3−21x−35=0whatisthevalueof(a−b)(c−a)(b−c)?
Commented by MJS_new last updated on 22/Jan/21
x^3 +px+q=0 Cardano gives u=−(q/2)+(√((q^2 /4)+(p^3 /(27))))∧v=−(q/2)−(√((q^2 /4)+(p^3 /(27)))) [and for this purpose they are allowed to be complex] the roots then are [ω=−(1/2)+((√3)/2)i] a=u^(1/3) +v^(1/3) b=ωu^(1/3) +w^2 v^(1/3) c=ω^2 u^(1/3) +ωv^(1/3) (a−b)(b−c)(c−a)= [knowing that ω^3 =1, ω^(3n+k) =ω^k ] ... =3(ω^2 −ω)(u−v)= =−3(√3)(u−v)i= =−i(√(4p^3 +27q^2 )) with p=−21∧q=−35 we get 63
x3+px+q=0Cardanogivesu=−q2+q24+p327∧v=−q2−q24+p327[andforthispurposetheyareallowedtobecomplex]therootsthenare[ω=−12+32i]a=u1/3+v1/3b=ωu1/3+w2v1/3c=ω2u1/3+ωv1/3(a−b)(b−c)(c−a)=[knowingthatω3=1,ω3n+k=ωk]…=3(ω2−ω)(u−v)==−33(u−v)i==−i4p3+27q2withp=−21∧q=−35weget63
Commented by bramlexs22 last updated on 15/Jan/21
correct sir. but how step to got it?
correctsir.buthowsteptogotit?
Commented by liberty last updated on 15/Jan/21
my way. let p(x) = (x−a)(x−b)(x−c) be polynomial has the roots are a,b and c so we have (x−a)(x−b)(x−c)=x^3 −21x−35 differentiating both sides w.r.t x gives (x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=3x^2 −21 putting x=a⇒3a^2 −21=(a−b)(a−c) putting x=b⇒3b^2 −21=(b−a)(b−c) putting x=c⇒3c^2 −21=(c−a)(c−b) multiply three equation we get (a−b)(a−c)(b−a)(b−c)(c−a)(c−b)=27(a^2 −7)(b^2 −7)(c^2 −7) [ (a−b)(b−c)(c−a) ]^2 = 27(a^2 −7)(b^2 −7)(c^2 −7) now consider the original equation x^3 −21x = 35 ; x(x^2 −21)=35 x^2 (x^2 −21)^2 =35^2 ; let y=x^2 −7 (y+7)(y+7−21)^2 =35^2 ⇒y^3 −21y^2 −147=0 whose roots are { ((a^2 −7 )),((b^2 −7 )),((c^2 −7)) :} then by Vieta′s rule we get (a^2 −7)(b^2 −7)(c^2 −7)=147 finally we find [ (a−b)(b−c)(c−a) ]^2 =27×147 ⇔ (a−b)(b−c)(c−a) = ± (√(3969)) = ± 63
myway.letp(x)=(x−a)(x−b)(x−c)bepolynomialhastherootsarea,bandcsowehave(x−a)(x−b)(x−c)=x3−21x−35differentiatingbothsidesw.r.txgives(x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=3x2−21puttingx=a⇒3a2−21=(a−b)(a−c)puttingx=b⇒3b2−21=(b−a)(b−c)puttingx=c⇒3c2−21=(c−a)(c−b)multiplythreeequationweget(a−b)(a−c)(b−a)(b−c)(c−a)(c−b)=27(a2−7)(b2−7)(c2−7)[(a−b)(b−c)(c−a)]2=27(a2−7)(b2−7)(c2−7)nowconsidertheoriginalequationx3−21x=35;x(x2−21)=35x2(x2−21)2=352;lety=x2−7(y+7)(y+7−21)2=352⇒y3−21y2−147=0whoserootsare{a2−7b2−7c2−7thenbyVieta′sruleweget(a2−7)(b2−7)(c2−7)=147finallywefind[(a−b)(b−c)(c−a)]2=27×147⇔(a−b)(b−c)(c−a)=±3969=±63―
Commented by MJS_new last updated on 15/Jan/21
I corrected and completed
Icorrectedandcompleted
Commented by bramlexs22 last updated on 15/Jan/21
waw...thank you
waw…thankyou
Commented by bemath last updated on 15/Jan/21
my method ⇔ x^3 −21x−35 = 0 a^3 −21a−35=0...(1) b^3 −21b−35=0...(2) (1)−(2)⇒a^3 −b^3 −21(a−b)=0 (a−b)(a^2 +ab+b^2 )−21(a−b)=0 (a−b)(a^2 +b^2 +ab−21)=0 ⇒(a−b)^2 +3ab−21=0 (a−b)^2 = 21−3ab similar to { (((c−a)^2 = 21−3ac)),(((b−c)^2 = 21−3bc)) :} so we find [(a−b)(c−a)(b−c)]^2 = (21−3ab)(21−3ac)(21−3bc) RHS ≡ −27(abc)^2 +189abc(a+b+c)− 1323(ab+ac+bc)+21^3 RHS≡ −27(35)^2 +189(0)−1323(−21)+21^3 RHS≡ 3969 finally we find (a−b)(c−a)(b−c) = ± (√(3969)) = ± 63
mymethod⇔x3−21x−35=0a3−21a−35=0…(1)b3−21b−35=0…(2)(1)−(2)⇒a3−b3−21(a−b)=0(a−b)(a2+ab+b2)−21(a−b)=0(a−b)(a2+b2+ab−21)=0⇒(a−b)2+3ab−21=0(a−b)2=21−3absimilarto{(c−a)2=21−3ac(b−c)2=21−3bcsowefind[(a−b)(c−a)(b−c)]2=(21−3ab)(21−3ac)(21−3bc)RHS≡−27(abc)2+189abc(a+b+c)−1323(ab+ac+bc)+213RHS≡−27(35)2+189(0)−1323(−21)+213RHS≡3969finallywefind(a−b)(c−a)(b−c)=±3969=±63

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