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Question Number 152663 by mr W last updated on 31/Aug/21

If x^3 -x+3=0 has the roots a, b and c. determine the monic polynomial with the roots a^5 , b^5 and c^5 . [Q152396]

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} -\mathrm{x}+\mathrm{3}=\mathrm{0}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{monic}\:\mathrm{polynomial}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\:\mathrm{a}^{\mathrm{5}} ,\:\mathrm{b}^{\mathrm{5}} \:\mathrm{and}\:\:\mathrm{c}^{\mathrm{5}} . \\ $$$$\left[{Q}\mathrm{152396}\right] \\ $$

Answered by mr W last updated on 30/Aug/21

a+b+c=0 ab+bc+ca=−1 abc=−3 let p_k =a^k +b^k +c^k p_1 =e_1 =0 p_2 =e_1 p_1 −2e_2 =−2×(−1)=2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =3(−3)=−9 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =−(−1)2=2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−(−1)(−9)+(−3)2=−15 i.e. a^5 +b^5 +c^5 =−15 let p_k =(ab)^k +(bc)^k +(ca)^k p_1 =e_1 =ab+bc+ca=−1 e_2 =ab^2 c+bc^2 a+ca^2 b=abc(a+b+c)=0 e_3 =ab×bc×ca=(abc)^2 =9 p_2 =e_1 p_1 −2e_2 =(−1)(−1)−2×0=1 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =(−1)1+3×9=26 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =(−1)26+9(−1)=−35 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =(−1)(−35)+9×1=44 i.e. a^5 b^5 +b^5 c^5 +c^5 a^5 =44 a^5 b^5 c^5 =(abc)^5 =(−3)^5 =−243 the equation with roots a^5 ,b^5 ,c^5 is x^3 +15x^2 +44x+243=0

$${a}+{b}+{c}=\mathrm{0} \\ $$$${ab}+{bc}+{ca}=−\mathrm{1} \\ $$$${abc}=−\mathrm{3} \\ $$$$ \\ $$$${let}\:{p}_{{k}} ={a}^{{k}} +{b}^{{k}} +{c}^{{k}} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{0} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =−\mathrm{2}×\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{3}\left(−\mathrm{3}\right)=−\mathrm{9} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =−\left(−\mathrm{1}\right)\mathrm{2}=\mathrm{2} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−\left(−\mathrm{1}\right)\left(−\mathrm{9}\right)+\left(−\mathrm{3}\right)\mathrm{2}=−\mathrm{15} \\ $$$${i}.{e}.\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} =−\mathrm{15} \\ $$$$ \\ $$$${let}\:{p}_{{k}} =\left({ab}\right)^{{k}} +\left({bc}\right)^{{k}} +\left({ca}\right)^{{k}} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} ={ab}+{bc}+{ca}=−\mathrm{1} \\ $$$${e}_{\mathrm{2}} ={ab}^{\mathrm{2}} {c}+{bc}^{\mathrm{2}} {a}+{ca}^{\mathrm{2}} {b}={abc}\left({a}+{b}+{c}\right)=\mathrm{0} \\ $$$${e}_{\mathrm{3}} ={ab}×{bc}×{ca}=\left({abc}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)−\mathrm{2}×\mathrm{0}=\mathrm{1} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\left(−\mathrm{1}\right)\mathrm{1}+\mathrm{3}×\mathrm{9}=\mathrm{26} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\left(−\mathrm{1}\right)\mathrm{26}+\mathrm{9}\left(−\mathrm{1}\right)=−\mathrm{35} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =\left(−\mathrm{1}\right)\left(−\mathrm{35}\right)+\mathrm{9}×\mathrm{1}=\mathrm{44} \\ $$$${i}.{e}.\:{a}^{\mathrm{5}} {b}^{\mathrm{5}} +{b}^{\mathrm{5}} {c}^{\mathrm{5}} +{c}^{\mathrm{5}} {a}^{\mathrm{5}} =\mathrm{44} \\ $$$$ \\ $$$${a}^{\mathrm{5}} {b}^{\mathrm{5}} {c}^{\mathrm{5}} =\left({abc}\right)^{\mathrm{5}} =\left(−\mathrm{3}\right)^{\mathrm{5}} =−\mathrm{243} \\ $$$$ \\ $$$${the}\:{equation}\:{with}\:{roots}\:{a}^{\mathrm{5}} ,{b}^{\mathrm{5}} ,{c}^{\mathrm{5}} \:{is} \\ $$$${x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{44}{x}+\mathrm{243}=\mathrm{0} \\ $$

Commented by Tawa11 last updated on 30/Aug/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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