Question-163437

Question Number 163437 by cortano1 last updated on 07/Jan/22

Commented by mr W last updated on 07/Jan/22

m^3 +n^3 +p^3 =(((13))^(1/3) )^3 +(((53))^(1/3) )^3 +(((103))^(1/3) )^3 −3×(1/3) =13+53+103−1=168

$${m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{13}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{53}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{103}}\right)^{\mathrm{3}} −\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\mathrm{13}+\mathrm{53}+\mathrm{103}−\mathrm{1}=\mathrm{168} \\ $$

Commented by cortano1 last updated on 07/Jan/22

$${what}\:{the}\:{formula}? \\ $$

Answered by mr W last updated on 07/Jan/22

say m,n,p are the roots of eqn. (x−α)(x−β)(x−γ)=δ, i.e. x^3 −(α+β+γ)x^2 +(αβ+βγ+γα)x+(αβγ−δ)=0 then we have m+n+p=α+β+γ mn+np+pm=αβ+βγ+γα mnp=αβγ−δ (m+n+p)^2 =m^2 +n^2 +p^2 +2(mn+np+pm) (α+β+γ)^2 =α^2 +β^2 +γ^2 +2(αβ+βγ+γα) ⇒m^2 +n^2 +p^2 =α^2 +β^2 +γ^2 ⇒(m^2 +n^2 +p^2 )(m+n+p)=(α^2 +β^2 +γ^2 )(α+α+γ) ⇒m^3 +n^3 +p^3 +(m+n+p)(mn+np+pm)−3mnp=α^3 +β^3 +β^3 +(α+β+γ)(αβ+βγ+γα)−3αβγ ⇒m^3 +n^3 +p^3 −3mnp=α^3 +β^3 +β^3 −3αβγ ⇒m^3 +n^3 +p^3 −3(αβγ−δ)=α^3 +β^3 +β^3 −3αβγ ⇒m^3 +n^3 +p^3 =α^3 +β^3 +β^3 −3δ with α=((13))^(1/3) , β=((53))^(1/3) , γ=((103))^(1/3) , δ=(1/3) ⇒m^3 +n^3 +p^3 =(((13))^(1/3) )^3 +(((53))^(1/3) )^3 +(((103))^(1/3) )^3 −3×(1/3) ⇒m^3 +n^3 +p^3 =13+53+103 −1=168 ■

$${say}\:{m},{n},{p}\:{are}\:{the}\:{roots}\:{of}\:{eqn}. \\ $$$$\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)=\delta,\:{i}.{e}. \\ $$$${x}^{\mathrm{3}} −\left(\alpha+\beta+\gamma\right){x}^{\mathrm{2}} +\left(\alpha\beta+\beta\gamma+\gamma\alpha\right){x}+\left(\alpha\beta\gamma−\delta\right)=\mathrm{0} \\ $$$${then}\:{we}\:{have} \\ $$$${m}+{n}+{p}=\alpha+\beta+\gamma \\ $$$${mn}+{np}+{pm}=\alpha\beta+\beta\gamma+\gamma\alpha \\ $$$${mnp}=\alpha\beta\gamma−\delta \\ $$$$\left({m}+{n}+{p}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} +{n}^{\mathrm{2}} +{p}^{\mathrm{2}} +\mathrm{2}\left({mn}+{np}+{pm}\right) \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{2}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right) \\ $$$$\Rightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} +{p}^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \\ $$$$\Rightarrow\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)\left({m}+{n}+{p}\right)=\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)\left(\alpha+\alpha+\gamma\right) \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} +\left({m}+{n}+{p}\right)\left({mn}+{np}+{pm}\right)−\mathrm{3}{mnp}=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\beta^{\mathrm{3}} +\left(\alpha+\beta+\gamma\right)\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)−\mathrm{3}\alpha\beta\gamma \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} −\mathrm{3}{mnp}=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\beta^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} −\mathrm{3}\left(\alpha\beta\gamma−\delta\right)=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\beta^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\beta^{\mathrm{3}} −\mathrm{3}\delta \\ $$$${with}\:\alpha=\sqrt[{\mathrm{3}}]{\mathrm{13}},\:\beta=\sqrt[{\mathrm{3}}]{\mathrm{53}},\:\gamma=\sqrt[{\mathrm{3}}]{\mathrm{103}},\:\delta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{13}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{53}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{103}}\right)^{\mathrm{3}} \:−\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\mathrm{13}+\mathrm{53}+\mathrm{103}\:−\mathrm{1}=\mathrm{168}\:\blacksquare \\ $$

Commented by Tawa11 last updated on 07/Jan/22