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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
$$\mathcal{D}{etermine}\:{a}\:,\:{b}\:,\:{c}\:{in}\:{terms}\:{of}\:\alpha\:,\:\beta\:,\:\gamma. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}+{c}=\gamma\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{bc}+{a}=\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ca}+{b}=\beta \\ $$
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 ......?
$${if}\:\alpha=\beta=\gamma=\mathrm{0},\:\Rightarrow{a}={b}={c}=\mathrm{0}\:{or}\:−\mathrm{1} \\ $$$${if}\:\alpha=\beta=\gamma=\mathrm{2},\:\Rightarrow{a}={b}={c}=\mathrm{1}\:{or}\:−\mathrm{2} \\ $$$${if}\:\alpha=\beta=\gamma={k},\:\Rightarrow{a}={b}={c}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{k}}}{\mathrm{2}} \\ $$$${otherwise}: \\ $$$${a}^{\mathrm{2}} {b}+{ac}={a}\gamma \\ $$$$−\left({iii}\right): \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{1}\right){b}={a}\gamma−\beta \\ $$$$\Rightarrow{b}=\frac{{a}\gamma−\beta}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${a}^{\mathrm{2}} {c}+{ab}={a}\beta \\ $$$$−\left({i}\right): \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{1}\right){c}={a}\beta−\gamma \\ $$$${c}=\frac{{a}\beta−\gamma}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${put}\:{into}\:\left({ii}\right): \\ $$$$\frac{\left({a}\beta−\gamma\right)\left({a}\gamma−\beta\right)}{\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+{a}=\alpha \\ $$$$\beta\gamma{a}^{\mathrm{2}} −\left(\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right){a}+\beta\gamma+{a}^{\mathrm{5}} −\mathrm{2}{a}^{\mathrm{3}} +{a}−\alpha{a}^{\mathrm{4}} +\mathrm{2}\alpha{a}^{\mathrm{2}} −\alpha=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{5}} −\alpha{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{3}} +\left(\mathrm{2}\alpha+\beta\gamma\right){a}^{\mathrm{2}} −\left(\mathrm{1}+\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right){a}−\left(\alpha−\beta\gamma\right)=\mathrm{0} \\ $$$$……? \\ $$
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 .
$$\mathrm{c}=\gamma−\mathrm{ab} \\ $$$$\mathrm{b}\left(\gamma−\mathrm{ab}\right)+\mathrm{a}=\alpha \\ $$$$\mathrm{a}\left(\gamma−\mathrm{ab}\right)+\mathrm{b}=\beta \\ $$$$\mathrm{a}+\mathrm{b}=\frac{\alpha+\beta}{\mathrm{1}+\gamma−\mathrm{ab}} \\ $$$$\mathrm{a}−\mathrm{b}=\frac{\alpha−\beta}{\mathrm{1}−\gamma+\mathrm{ab}} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{4ab} \\ $$$$\Rightarrow\:\:\frac{\left(\alpha+\beta\right)^{\mathrm{2}} }{\left(\mathrm{1}+\gamma−\mathrm{ab}\right)^{\mathrm{2}} }+\frac{\left(\alpha−\beta\right)^{\mathrm{2}} }{\left(\mathrm{1}−\gamma+\mathrm{ab}\right)^{\mathrm{2}} }=\mathrm{4ab} \\ $$$$\mathrm{let}\:\:\mathrm{ab}=\mathrm{x}\:,\:\alpha+\beta=\mathrm{s}\:,\:\alpha−\beta=\mathrm{d} \\ $$$$\mathrm{s}^{\mathrm{2}} \left(\mathrm{1}−\gamma+\mathrm{x}\right)^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \left(\mathrm{1}+\gamma−\mathrm{x}\right)^{\mathrm{2}} =\mathrm{4x}\left(\mathrm{1}+\gamma−\mathrm{x}\right)^{\mathrm{2}} \left(\mathrm{1}−\gamma+\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{quintic}\:\:,\:\mathrm{ab}=\mathrm{x} \\ $$$$\mathrm{Now}\:\:\:\mathrm{2a}=\frac{\mathrm{s}}{\mathrm{1}+\gamma−\mathrm{x}}+\frac{\mathrm{d}}{\mathrm{1}−\gamma+\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2b}=\frac{\mathrm{s}}{\mathrm{1}+\gamma−\mathrm{x}}−\frac{\mathrm{d}}{\mathrm{1}−\gamma+\mathrm{x}}\:\:\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\gamma−\mathrm{x}\:. \\ $$

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