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a-b-c-R-a-b-c-1-a-2-b-2-c-2-1-a-10-b-10-c-10-1-a-4-b-4-c-4-

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Question Number 212224 by RojaTaniya last updated on 07/Oct/24
a,b,c∈R a+b+c=1, a^2 +b^2 +c^2 =1 a^(10) +b^(10) +c^(10) =1, a^4 +b^4 +c^4 =?
$$\:{a},{b},{c}\in{R} \\ $$$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$
Answered by mr W last updated on 07/Oct/24
1^2 =1+2(ab+bc+ca) ⇒ab+bc+ca=0 0^2 =a^2 b^2 +b^2 c^2 +c^2 a^2 +2abc(a+b+c) ⇒a^2 b^2 +b^2 c^2 +c^2 a^2 =−2abc=−2k (−2k)^2 =a^4 b^4 +b^4 c^4 +c^4 a^4 +2a^2 b^2 c^2 (a^2 +b^2 +c^2 ) ⇒a^4 b^4 +b^4 c^4 +c^4 a^4 =2k^2 1^2 =a^4 +b^4 +c^4 +2(a^2 b^2 +b^2 c^2 +c^2 a^2 ) ⇒a^4 +b^4 +c^4 =1+4k a^6 +b^6 +c^6 +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a^2 +b^2 +c^2 )−3a^2 b^2 c^2 =(1+4k)×1 ⇒a^6 +b^6 +c^6 =1+6k+3k^2 (1+4k)^2 =a^8 +b^8 +c^8 +2(a^4 b^4 +b^4 c^4 +c^4 a^4 ) ⇒a^8 +b^8 +c^8 =1+8k+12k^2 a^(10) +b^(10) +c^(10) +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a^6 +b^6 +c^6 )−a^2 b^2 c^2 (a^4 +b^4 +c^4 )=(1+8k+12k^2 )×1 1+(−2k)(1+6k+3k^2 )−k^2 (1+4k)=(1+8k+12k^2 )×1 ⇒k(2k+1)(k+2)=0 ⇒k=0, −(1/2), −2 ⇒a^4 +b^4 +c^4 =1+4k=1, −1, −7 if a,b,c ∈R: a^4 +b^4 +c^4 =1
$$\mathrm{1}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\left({ab}+{bc}+{ca}\right)\: \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{0} \\ $$$$\mathrm{0}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}{abc}\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} =−\mathrm{2}{abc}=−\mathrm{2}{k} \\ $$$$\left(−\mathrm{2}{k}\right)^{\mathrm{2}} ={a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} =\mathrm{2}{k}^{\mathrm{2}} \\ $$$$\mathrm{1}^{\mathrm{2}} ={a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1}+\mathrm{4}{k} \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} +\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{4}{k}\right)×\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\mathrm{4}{k}\right)^{\mathrm{2}} ={a}^{\mathrm{8}} +{b}^{\mathrm{8}} +{c}^{\mathrm{8}} +\mathrm{2}\left({a}^{\mathrm{4}} {b}^{\mathrm{4}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} +{c}^{\mathrm{4}} {a}^{\mathrm{4}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{8}} +{b}^{\mathrm{8}} +{c}^{\mathrm{8}} =\mathrm{1}+\mathrm{8}{k}+\mathrm{12}{k}^{\mathrm{2}} \\ $$$${a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} +\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)=\left(\mathrm{1}+\mathrm{8}{k}+\mathrm{12}{k}^{\mathrm{2}} \right)×\mathrm{1} \\ $$$$\mathrm{1}+\left(−\mathrm{2}{k}\right)\left(\mathrm{1}+\mathrm{6}{k}+\mathrm{3}{k}^{\mathrm{2}} \right)−{k}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4}{k}\right)=\left(\mathrm{1}+\mathrm{8}{k}+\mathrm{12}{k}^{\mathrm{2}} \right)×\mathrm{1} \\ $$$$\Rightarrow{k}\left(\mathrm{2}{k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{0},\:−\frac{\mathrm{1}}{\mathrm{2}},\:−\mathrm{2} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1}+\mathrm{4}{k}=\mathrm{1},\:−\mathrm{1},\:−\mathrm{7} \\ $$$${if}\:{a},{b},{c}\:\in{R}:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{1} \\ $$

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