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Question Number 170836 by balirampatel last updated on 01/Jun/22

x^3 − 2x^2 − 5x + 6 = 0 α^3 + β^3 + γ^3 = ?

$${x}^{\mathrm{3}} \:−\:\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \:+\:\gamma^{\mathrm{3}} \:=\:? \\ $$

Commented by cortano1 last updated on 01/Jun/22

(x−1)(x^2 −x−6)=0 (x−1)(x−3)(x+2)=0 α^3 +β^3 +γ^3 = 1^3 +3^3 +(−2)^3 =20

$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \:=\:\mathrm{1}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\left(−\mathrm{2}\right)^{\mathrm{3}} =\mathrm{20} \\ $$

Answered by pablo1234523 last updated on 01/Jun/22

α^3 +β^3 +γ^3 =(α+β+γ)(α^2 +β^2 +γ^2 −αβ−βγ−γα)+3αβγ =(α+β+γ)[(α+β+γ)^2 −3(αβ+βγ+γα)]+3αβγ =(2)[2^2 −3×(−5)]+3×(−6) =20

$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:=\left(\alpha+\beta+\gamma\right)\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} −\alpha\beta−\beta\gamma−\gamma\alpha\right)+\mathrm{3}\alpha\beta\gamma \\ $$$$\:\:\:\:\:\:\:\:=\left(\alpha+\beta+\gamma\right)\left[\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{3}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)\right]+\mathrm{3}\alpha\beta\gamma \\ $$$$\:\:\:\:\:\:\:\:=\left(\mathrm{2}\right)\left[\mathrm{2}^{\mathrm{2}} −\mathrm{3}×\left(−\mathrm{5}\right)\right]+\mathrm{3}×\left(−\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{20} \\ $$

Answered by som(math1967) last updated on 01/Jun/22

α+β+γ=2 (αβ+βγ+γα)=−5 αβγ=−6 α^3 +β^3 +γ^3 =(α+β+γ)^3 −3(α+β)(β+γ)(γ+α) =2^3 −3{(α+β+γ)(αβ+βγ+γα)−αβγ}★ =8−3{2×−5−(−6)} =8−3(−10+6) =8+12=20 ★(𝛂+𝛃)(𝛃+𝛄)(𝛄+𝛂) =(𝛂+𝛃+𝛄)(𝛂𝛃+𝛃𝛄+𝛄𝛂)−𝛂𝛃𝛄

$$\alpha+\beta+\gamma=\mathrm{2}\:\:\:\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)=−\mathrm{5} \\ $$$$\alpha\beta\gamma=−\mathrm{6} \\ $$$$\:\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \\ $$$$=\left(\alpha+\beta+\gamma\right)^{\mathrm{3}} −\mathrm{3}\left(\alpha+\beta\right)\left(\beta+\gamma\right)\left(\gamma+\alpha\right) \\ $$$$=\mathrm{2}^{\mathrm{3}} −\mathrm{3}\left\{\left(\alpha+\beta+\gamma\right)\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)−\alpha\beta\gamma\right\}\bigstar \\ $$$$=\mathrm{8}−\mathrm{3}\left\{\mathrm{2}×−\mathrm{5}−\left(−\mathrm{6}\right)\right\} \\ $$$$=\mathrm{8}−\mathrm{3}\left(−\mathrm{10}+\mathrm{6}\right) \\ $$$$=\mathrm{8}+\mathrm{12}=\mathrm{20} \\ $$$$\bigstar\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)\left(\boldsymbol{\beta}+\boldsymbol{\gamma}\right)\left(\boldsymbol{\gamma}+\boldsymbol{\alpha}\right) \\ $$$$\:=\left(\boldsymbol{\alpha}+\boldsymbol{\beta}+\boldsymbol{\gamma}\right)\left(\boldsymbol{\alpha\beta}+\boldsymbol{\beta\gamma}+\boldsymbol{\gamma\alpha}\right)−\boldsymbol{\alpha\beta\gamma} \\ $$

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