Question Number 220468 by Rojarani last updated on 13/May/25 | ||
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$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$ | ||
Answered by mr W last updated on 15/May/25 | ||
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$${say}\:{abc}={k} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{0} \\ $$$$\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}{abc}\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} =−\mathrm{2}{k} \\ $$$${a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} +\mathrm{2}\left({abc}\right)^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{4}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} =\mathrm{2}{k}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1}+\mathrm{4}{k} \\ $$$$\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{1}+\mathrm{4}{k} \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\mathrm{3}\left({abc}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{4}{k} \\ $$$$\Rightarrow{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} \right)\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)=\left(\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{k}\right) \\ $$$${a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} \right)−\mathrm{3}\left({abc}\right)^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{k}\right) \\ $$$$\mathrm{1}+\mathrm{2}{k}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{k}\right) \\ $$$${k}\left(\mathrm{6}{k}^{\mathrm{2}} +\mathrm{14}{k}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{0}\:{or}\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{19}}}{\mathrm{6}} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1}+\mathrm{4}{k}=\mathrm{1}\:{or}\:\frac{−\mathrm{11}\pm\mathrm{2}\sqrt{\mathrm{19}}}{\mathrm{3}} \\ $$$${if}\:{a},\:{b},\:{c}\:\in\:{R},\:{then}\:\frac{−\mathrm{11}\pm\mathrm{2}\sqrt{\mathrm{19}}}{\mathrm{3}}<\mathrm{0}\: \\ $$$${must}\:{be}\:{rejected}. \\ $$ | ||
Commented by Rojarani last updated on 13/May/25 | ||
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$$\:{Sir},\:{excellent}.\:{Thanks}. \\ $$ | ||
Commented by Rojarani last updated on 15/May/25 | ||
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$$\:{Sir},\:\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\:={a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right) \\ $$$$\:{kindly}\:{check}\:{the}\:{identity}. \\ $$ | ||
Commented by mr W last updated on 15/May/25 | ||
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$${fixed}!\:{thanks}\:{for}\:{checking}! \\ $$ | ||
Commented by Rojarani last updated on 15/May/25 | ||
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$${Sir},\:{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \\ $$ | ||
Answered by MrGaster last updated on 13/May/25 | ||
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$$\begin{cases}{{a}+{b}+{c}=\mathrm{1}}\\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}}\\{{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1}}\end{cases} \\ $$$${S}_{\mathrm{1}} ={a}+{b}+{c}=\mathrm{1} \\ $$$${S}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$${S}_{\mathrm{10}} ={a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{symmetric}\:\mathrm{polynomial}\:\mathrm{relations}: \\ $$$${S}_{\mathrm{2}} ={S}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}\left({ab}−{bc}+{ca}\right) \\ $$$$\mathrm{1}=\mathrm{1}^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${ab}+{bc}+{ca}=\mathrm{0} \\ $$$$\mathrm{Recursive}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{higher}-\mathrm{order}\:\mathrm{symmetric}\:\mathrm{sums}: \\ $$$${S}_{\mathrm{3}} ={S}_{\mathrm{1}} \centerdot{S}_{\mathrm{2}} −\left({ab}+{bc}+{ca}\right)\centerdot{S}_{\mathrm{1}} +\mathrm{3}{abc}=\mathrm{1}+\mathrm{3}{abc} \\ $$$${S}_{\mathrm{4}} ={S}_{\mathrm{1}} \centerdot{S}_{\mathrm{3}} −\left({ab}+{bc}+{ca}\right)\centerdot{S}_{\mathrm{2}} +{abc}\centerdot{S}_{\mathrm{1}} =\mathrm{1}+\mathrm{4}{abc} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:{S}_{\mathrm{1}} =\mathrm{1}\:\mathrm{and}\:{S}_{\mathrm{2}} =\mathrm{1},\mathrm{so}: \\ $$$${S}_{\mathrm{3}} =\mathrm{1}+\mathrm{3}{abc}=\mathrm{1}\Rightarrow{abc}=\mathrm{0} \\ $$$$\mathrm{Now},\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:{S}_{\mathrm{4}} : \\ $$$${S}_{\mathrm{4}} =\mathrm{1}+\mathrm{4}{abc}=\mathrm{1}+\mathrm{4}\centerdot\mathrm{0}=\mathrm{1} \\ $$$$\mathrm{Therefore},\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \mathrm{is}\:\mathrm{1} \\ $$ | ||
Commented by Rojarani last updated on 13/May/25 | ||
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$$\:{Sir},\:{how}\:{decide}\:{that}\:{S}_{\mathrm{3}} =\mathrm{1}? \\ $$ | ||
Answered by Ghisom last updated on 13/May/25 | ||
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$${a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =−\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}=\mathrm{0}\:\Rightarrow\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =−\mathrm{7} \\ $$$$ \\ $$$$\left[{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\alpha=\mathrm{0}\:\Rightarrow\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1}−\mathrm{4}\alpha\right] \\ $$ | ||
Commented by Rojarani last updated on 14/May/25 | ||
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$$\:{Sir}\:{how}\:{can}\:{find}\:\:{abc}=? \\ $$ | ||