Question Number 169610 by mathlove last updated on 04/May/22
Answered by som(math1967) last updated on 04/May/22
![(α^2 /(α^2 +αβ+αγ))+(β^2 /(β^2 +αβ+βγ))+(γ^2 /(γ^2 +βγ+γα)) ★ =(α^2 /(α(α+β+γ))) +(β^2 /(β(α+β+γ)))+(γ^2 /(γ(α+β+γ))) =(((α+β+γ))/((α+β+γ)))=1 [α+β+γ≠0] ★αβ+βγ+γα=0 ∴−βγ=αβ+αγ −γα=αβ+βγ −αβ=βγ+γα](https://www.tinkutara.com/question/Q169611.png)
$$\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\alpha\beta+\alpha\gamma}+\frac{\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\alpha\beta+\beta\gamma}+\frac{\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\beta\gamma+\gamma\alpha}\:\bigstar \\ $$$$=\frac{\alpha^{\mathrm{2}} }{\alpha\left(\alpha+\beta+\gamma\right)}\:+\frac{\beta^{\mathrm{2}} }{\beta\left(\alpha+\beta+\gamma\right)}+\frac{\gamma^{\mathrm{2}} }{\gamma\left(\alpha+\beta+\gamma\right)} \\ $$$$=\frac{\left(\alpha+\beta+\gamma\right)}{\left(\alpha+\beta+\gamma\right)}=\mathrm{1}\:\left[\alpha+\beta+\gamma\neq\mathrm{0}\right] \\ $$$$\bigstar\alpha\beta+\beta\gamma+\gamma\alpha=\mathrm{0} \\ $$$$\therefore−\beta\gamma=\alpha\beta+\alpha\gamma \\ $$$$\:−\gamma\alpha=\alpha\beta+\beta\gamma \\ $$$$−\alpha\beta=\beta\gamma+\gamma\alpha \\ $$