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Question Number 135467 by benjo_mathlover last updated on 13/Mar/21
Answered by EDWIN88 last updated on 13/Mar/21
![determinant (((α β γ)),((β γ α)),((γ α β)))= α(βγ−α^2 )−β(β^2 −αγ)+γ(αβ−γ^2 ) =αβγ−α^3 −β^3 +αβγ+αβγ−γ^3 =3αβγ−(α^3 +β^3 +γ^3 )...(i) = −3b−[ −a^3 −3b ] = a^3 consider : x^3 +ax^2 +b=0 { (α),(β),(γ) :} ⇒α^3 = −aα^2 −b ⇒β^3 = −aβ^2 −b ⇒γ^3 = −aγ^2 −b ______________ + :α^3 +β^3 +γ^3 =−a(α^2 +β^2 +γ^2 )−3b =−a[(α+β+γ)^2 −2(αβ+αγ+βγ)]−3b =−a[ (−a)^2 −2(0)]−3b=−a^3 −3b](Q135476.png)
$$\begin{vmatrix}{\alpha\:\:\:\:\:\beta\:\:\:\:\:\:\gamma}\\{\beta\:\:\:\:\:\gamma\:\:\:\:\:\:\alpha}\\{\gamma\:\:\:\:\:\alpha\:\:\:\:\:\beta}\end{vmatrix}=\:\alpha\left(\beta\gamma−\alpha^{\mathrm{2}} \right)−\beta\left(\beta^{\mathrm{2}} −\alpha\gamma\right)+\gamma\left(\alpha\beta−\gamma^{\mathrm{2}} \right) \\ $$$$=\alpha\beta\gamma−\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} +\alpha\beta\gamma+\alpha\beta\gamma−\gamma^{\mathrm{3}} \\ $$$$=\mathrm{3}\alpha\beta\gamma−\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)...\left(\mathrm{i}\right) \\ $$$$=\:−\mathrm{3b}−\left[\:−\mathrm{a}^{\mathrm{3}} −\mathrm{3b}\:\right] \\ $$$$=\:\mathrm{a}^{\mathrm{3}} \: \\ $$$$\mathrm{consider}\::\:\mathrm{x}^{\mathrm{3}} +\mathrm{ax}^{\mathrm{2}} +\mathrm{b}=\mathrm{0\begin{cases}{\alpha}\\{\beta}\\{\gamma}\end{cases}} \\ $$$$\Rightarrow\alpha^{\mathrm{3}} \:=\:−\mathrm{a}\alpha^{\mathrm{2}} −\mathrm{b} \\ $$$$\Rightarrow\beta^{\mathrm{3}} =\:−\mathrm{a}\beta^{\mathrm{2}} −\mathrm{b} \\ $$$$\Rightarrow\gamma^{\mathrm{3}} =\:−\mathrm{a}\gamma^{\mathrm{2}} −\mathrm{b} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:+ \\ $$$$:\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} =−\mathrm{a}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)−\mathrm{3b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{a}\left[\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\alpha\gamma+\beta\gamma\right)\right]−\mathrm{3b} \\ $$$$\:\:\:\:=−\mathrm{a}\left[\:\left(−\mathrm{a}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{0}\right)\right]−\mathrm{3b}=−\mathrm{a}^{\mathrm{3}} −\mathrm{3b} \\ $$
Commented by mr W last updated on 13/Mar/21
![=−a[ (−a)^2 −2(0)]−3b=−a^3 −3b ⇒answer c](Q135481.png)
$$\:\:\:\:=−\mathrm{a}\left[\:\left(−{a}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{0}\right)\right]−\mathrm{3b}=−\mathrm{a}^{\mathrm{3}} −\mathrm{3b} \\ $$$$\Rightarrow{answer}\:{c} \\ $$
Commented by EDWIN88 last updated on 13/Mar/21
$$\mathrm{hahaha} \\ $$