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Question Number 143602 by meetbhavsar25 last updated on 16/Jun/21
Commented by TheHoneyCat last updated on 16/Jun/21
Are you asking for Π_(a∈C) g(a)^(val_f (a)) or z such that ∀(a_i )_(i∈[∣1,6∣]) Π_(i=1) ^6 g(a_i )=z ¿ (that is to say, is each a_i one sigle root of f or can they all be the same for instance)
Areyouaskingfor∏a∈Cg(a)valf(a)orzsuchthat∀(ai)i∈[∣1,6∣]∏6i=1g(ai)=z¿(thatistosay,iseachaionesiglerootofforcantheyallbethesameforinstance)
Answered by mr W last updated on 16/Jun/21
f(x)=(x^3 +2)(x^3 −4)=0 ⇒α_1 ^3 ,α_2 ^3 ,α_3 ^3 =−2 ⇒α_(1,2,) =−(2)^(1/3) ,−(2)^(1/3) ω,−(2)^(1/3) ω^2 ⇒α_4 ^3 ,α_5 ^3 ,α_6 ^3 =4 ⇒α_(4,5,6) =(4)^(1/3) ,(4)^(1/3) ω,(4)^(1/3) ω^2 g(x)=(x+1+i(√3))(x+1−i(√3)) g(x)=(x−2ω)(x−2ω^2 ) Π_(k=1) ^6 g(α_k )=Π_(k=1) ^6 (α_k −2ω)(α_k −2ω^2 ) =(−(2)^(1/3) −2ω)(−(2)^(1/3) −2ω^2 ) ×(−(2)^(1/3) ω−2ω)(−(2)^(1/3) ω−2ω^2 ) ×(−(2)^(1/3) ω^2 −2ω)(−(2)^(1/3) ω^2 −2ω^2 ) ×((4)^(1/3) −2ω)((4)^(1/3) −2ω^2 ) ×((4)^(1/3) ω−2ω)((4)^(1/3) ω−2ω^2 ) ×((4)^(1/3) ω^2 −2ω)((4)^(1/3) ω^2 −2ω^2 ) =((2)^(1/3) +2ω)((2)^(1/3) ω+2)((2)^(1/3) +2)^2 ((2)^(1/3) +2ω)((2)^(1/3) ω+2) ×((4)^(1/3) −2ω)((4)^(1/3) ω−2)((4)^(1/3) −2)^2 ((4)^(1/3) −2ω)((4)^(1/3) ω−2)ω^2 =((2)^(1/3) +2ω)^2 ((2)^(1/3) ω+2)^2 ((2)^(1/3) +2)^2 ×((4)^(1/3) −2ω)^2 ((4)^(1/3) ω−2)^2 ((4)^(1/3) −2)^2 ω^2 =1600
f(x)=(x3+2)(x3−4)=0⇒α13,α23,α33=−2⇒α1,2,=−23,−23ω,−23ω2⇒α43,α53,α63=4⇒α4,5,6=43,43ω,43ω2g(x)=(x+1+i3)(x+1−i3)g(x)=(x−2ω)(x−2ω2)∏6k=1g(αk)=∏6k=1(αk−2ω)(αk−2ω2)=(−23−2ω)(−23−2ω2)×(−23ω−2ω)(−23ω−2ω2)×(−23ω2−2ω)(−23ω2−2ω2)×(43−2ω)(43−2ω2)×(43ω−2ω)(43ω−2ω2)×(43ω2−2ω)(43ω2−2ω2)=(23+2ω)(23ω+2)(23+2)2(23+2ω)(23ω+2)×(43−2ω)(43ω−2)(43−2)2(43−2ω)(43ω−2)ω2=(23+2ω)2(23ω+2)2(23+2)2×(43−2ω)2(43ω−2)2(43−2)2ω2=1600
Answered by ajfour last updated on 16/Jun/21
x^6 −2x^3 −8=0 and x^2 =g−2x−4 (g−2x−4)^3 −2x(g−2x−4)=8 (g−4)^3 −8x(g−2x−4) −6(g−4)^2 x+12(g−4)(g−2x−4) −2(g−4)x+4(g−2x−4)−8=0 ⇒ (g−4)^3 +16(g−2x−4) −8x(g−4)−6(g−4)^2 x +12(g−4)^2 −24(g−4)x −2(g−4)x+4(g−4)−8x−8 =0 ⇒ x=(((g−4)^3 +12(g−4)^2 +20(g−4)−8)/(6(g−4)^2 +34(g−4)+40)) ⇒ but (x+1)^2 =g−3 ⇒ {(((g−4)^3 +18(g−4)^2 +54(g−4)+32)/(6(g−4)^2 +34(g−4)+40))}^2 =g−3 Πg_i ={(−4)^3 +18(−4)^2 +54(−4)+32}^2 −(−3){6(−4)^2 +34(−4)+40}^2 =16(−16+72−54+8)^2 +48(24−34+10)^2 = 1600 ★ yes sir!
x6−2x3−8=0andx2=g−2x−4(g−2x−4)3−2x(g−2x−4)=8(g−4)3−8x(g−2x−4)−6(g−4)2x+12(g−4)(g−2x−4)−2(g−4)x+4(g−2x−4)−8=0⇒(g−4)3+16(g−2x−4)−8x(g−4)−6(g−4)2x+12(g−4)2−24(g−4)x−2(g−4)x+4(g−4)−8x−8=0⇒x=(g−4)3+12(g−4)2+20(g−4)−86(g−4)2+34(g−4)+40⇒but(x+1)2=g−3⇒{(g−4)3+18(g−4)2+54(g−4)+326(g−4)2+34(g−4)+40}2=g−3Πgi={(−4)3+18(−4)2+54(−4)+32}2−(−3){6(−4)2+34(−4)+40}2=16(−16+72−54+8)2+48(24−34+10)2=1600★yessir!
Commented by mr W last updated on 16/Jun/21
i got 1600. but my method is not as smart.
igot1600.butmymethodisnotassmart.

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