Question Number 224728 by mr W last updated on 30/Sep/25
Commented by fantastic last updated on 30/Sep/25
$$\mathrm{170}?? \\ $$
Commented by mr W last updated on 30/Sep/25
$${yes} \\ $$
Answered by mr W last updated on 01/Oct/25
$${m},{n},{p}\:{are}\:{roots}\:{of} \\ $$$$\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)=\delta \\ $$$$\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)−\delta=\left({x}−{m}\right)\left({x}−{n}\right)\left({x}−{p}\right) \\ $$$${x}^{\mathrm{3}} −\left(\alpha+\beta+\gamma\right){x}^{\mathrm{2}} +\left(\alpha\beta+\beta\gamma+\gamma\alpha\right){x}−\left(\alpha\beta\gamma+\delta\right)= \\ $$$${x}^{\mathrm{3}} −\left({m}+{n}+{p}\right){x}^{\mathrm{2}} +\left({mn}+{np}+{pm}\right){x}−{mnp}=\mathrm{0} \\ $$$$\Rightarrow{m}+{n}+{p}=\alpha+\beta+\gamma \\ $$$$\Rightarrow{mn}+{np}+{pm}=\alpha\beta+\beta\gamma+\gamma\alpha \\ $$$$\Rightarrow{mnp}=\alpha\beta\gamma+\delta \\ $$$$\left({m}+{n}+{p}\right)^{\mathrm{3}} ={m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} −\mathrm{3}{mnp}+\mathrm{3}\left({m}+{n}+{p}\right)\left({mn}+{np}+{pm}\right) \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{3}} =\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma+\mathrm{3}\left(\alpha+\beta+\gamma\right)\left(\alpha\beta+\beta\gamma+\gamma\alpha\right) \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} −\mathrm{3}{mnp}=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} −\mathrm{3}\left(\alpha\beta\gamma+\delta\right)=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} +\mathrm{3}\delta \\ $$$$\:\:\:\:\:\:=\left(\sqrt[{\mathrm{3}}]{\mathrm{13}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{53}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{103}}\right)^{\mathrm{3}} +\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{170} \\ $$
Answered by Ghisom_ last updated on 30/Sep/25
$${u}^{\mathrm{3}} +{au}^{\mathrm{2}} +{bu}+{c}=\mathrm{0} \\ $$$${u}\in\left\{{j},\:{k},\:{l}\right\} \\ $$$$\Rightarrow \\ $$$${v}\in\left\{{j}^{\mathrm{3}} ,\:{k}^{\mathrm{3}} ,{l}^{\mathrm{3}} \right\}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$${v}^{\mathrm{3}} +\left({a}^{\mathrm{3}} −\mathrm{3}{ab}+\mathrm{3}{c}\right){v}^{\mathrm{2}} −\left(\mathrm{3}{abc}−{b}^{\mathrm{3}} −\mathrm{3}{c}^{\mathrm{2}} \right){v}+{c}^{\mathrm{3}} =\mathrm{0} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\left({u}−\alpha\right)\left({u}−\beta\right)\left({u}−\gamma\right)−\delta=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${a}=−\left(\alpha+\beta+\gamma\right)\wedge{b}=\alpha\beta+\beta\gamma+\gamma\alpha\wedge{c}=−\left(\alpha\beta\gamma+\delta\right) \\ $$$$\Rightarrow \\ $$$${u}^{\mathrm{3}} −\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} +\mathrm{3}\delta\right){u}^{\mathrm{2}} +{Qu}+{R}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{13}+\mathrm{53}+\mathrm{103}+\mathrm{1}=\mathrm{170} \\ $$
Commented by mr W last updated on 01/Oct/25
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