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Question Number 59581 by Rasheed.Sindhi last updated on 12/May/19

Determine a , b , c in terms of α , β , γ. ab+c=γ bc+a=α ca+b=β

Determinea,b,cintermsofα,β,γ.ab+c=γbc+a=αca+b=β

Answered by mr W last updated on 13/May/19

if α=β=γ=0, ⇒a=b=c=0 or −1 if α=β=γ=2, ⇒a=b=c=1 or −2 if α=β=γ=k, ⇒a=b=c=((−1±(√(1+4k)))/2) otherwise: a^2 b+ac=aγ −(iii): (a^2 −1)b=aγ−β ⇒b=((aγ−β)/(a^2 −1)) a^2 c+ab=aβ −(i): (a^2 −1)c=aβ−γ c=((aβ−γ)/(a^2 −1)) put into (ii): (((aβ−γ)(aγ−β))/((a^2 −1)^2 ))+a=α βγa^2 −(β^2 +γ^2 )a+βγ+a^5 −2a^3 +a−αa^4 +2αa^2 −α=0 ⇒a^5 −αa^4 −2a^3 +(2α+βγ)a^2 −(1+β^2 +γ^2 )a−(α−βγ)=0 ......?

ifα=β=γ=0,⇒a=b=c=0or−1ifα=β=γ=2,⇒a=b=c=1or−2ifα=β=γ=k,⇒a=b=c=−1±1+4k2otherwise:a2b+ac=aγ−(iii):(a2−1)b=aγ−β⇒b=aγ−βa2−1a2c+ab=aβ−(i):(a2−1)c=aβ−γc=aβ−γa2−1putinto(ii):(aβ−γ)(aγ−β)(a2−1)2+a=αβγa2−(β2+γ2)a+βγ+a5−2a3+a−αa4+2αa2−α=0⇒a5−αa4−2a3+(2α+βγ)a2−(1+β2+γ2)a−(α−βγ)=0......?

Answered by ajfour last updated on 12/May/19

c=γ−ab b(γ−ab)+a=α a(γ−ab)+b=β a+b=((α+β)/(1+γ−ab)) a−b=((α−β)/(1−γ+ab)) (a+b)^2 −(a−b)^2 =4ab ⇒ (((α+β)^2 )/((1+γ−ab)^2 ))+(((α−β)^2 )/((1−γ+ab)^2 ))=4ab let ab=x , α+β=s , α−β=d s^2 (1−γ+x)^2 +d^2 (1+γ−x)^2 =4x(1+γ−x)^2 (1−γ+x)^2 quintic , ab=x Now 2a=(s/(1+γ−x))+(d/(1−γ+x)) 2b=(s/(1+γ−x))−(d/(1−γ+x)) . c=γ−x .

c=γ−abb(γ−ab)+a=αa(γ−ab)+b=βa+b=α+β1+γ−aba−b=α−β1−γ+ab(a+b)2−(a−b)2=4ab⇒(α+β)2(1+γ−ab)2+(α−β)2(1−γ+ab)2=4abletab=x,α+β=s,α−β=ds2(1−γ+x)2+d2(1+γ−x)2=4x(1+γ−x)2(1−γ+x)2quintic,ab=xNow2a=s1+γ−x+d1−γ+x2b=s1+γ−x−d1−γ+x.c=γ−x.

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