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Question-215469

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Question Number 215469 by Abdullahrussell last updated on 08/Jan/25
Answered by alephnull last updated on 08/Jan/25
((1/2^2 )+(1/3^3 ))=((1/4)+(1/(27))) =(((27)/(108))+(4/(108))) the root is ((31)/(108)) sum of coeffecients = P(1) let Q(n)=108∙P(n) P(1)=((Q(1))/(108)) let Q(n) = a_6 n^6 +a_5 n^5 ...+a_1 n+a_0 P(1)=Q(1)/108 did you provide a specific polynomial? if not the sum depends on Q(n)
$$\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right) \\ $$$$=\left(\frac{\mathrm{27}}{\mathrm{108}}+\frac{\mathrm{4}}{\mathrm{108}}\right) \\ $$$${the}\:{root}\:{is}\:\frac{\mathrm{31}}{\mathrm{108}} \\ $$$$ \\ $$$${sum}\:{of}\:{coeffecients}\:=\:{P}\left(\mathrm{1}\right) \\ $$$${let}\:{Q}\left({n}\right)=\mathrm{108}\centerdot{P}\left({n}\right) \\ $$$${P}\left(\mathrm{1}\right)=\frac{{Q}\left(\mathrm{1}\right)}{\mathrm{108}} \\ $$$$ \\ $$$${let}\:{Q}\left({n}\right)\:=\:{a}_{\mathrm{6}} {n}^{\mathrm{6}} +{a}_{\mathrm{5}} {n}^{\mathrm{5}} …+{a}_{\mathrm{1}} {n}+{a}_{\mathrm{0}} \\ $$$${P}\left(\mathrm{1}\right)={Q}\left(\mathrm{1}\right)/\mathrm{108} \\ $$$${did}\:{you}\:{provide}\:{a}\:{specific}\:{polynomial}? \\ $$$${if}\:{not}\:{the}\:{sum}\:{depends}\:{on}\:{Q}\left({n}\right) \\ $$$$ \\ $$
Commented by Abdullahrussell last updated on 08/Jan/25
2^(1/2) +3^(1/3) =(√2) +(3)^(1/3)
$$\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} =\sqrt{\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3}}\: \\ $$
Answered by mr W last updated on 08/Jan/25
say P(x)=a_0 +a_1 x+a_2 x^2 +...+a_6 x^6 with a_6 =1 P(2^(1/2) +3^(1/3) )=Σ_(n=0) ^6 a_n (2^(1/2) +3^(1/3) )^n =0 (2^(1/2) +3^(1/3) )^1 =2^(1/2) +3^(1/3) (2^(1/2) +3^(1/3) )^2 =2+2×2^(1/2) ×3^(1/3) +3^(2/3) (2^(1/2) +3^(1/3) )^3 =3+2×2^(1/2) +6×3^(1/3) +3×2^(1/2) ×3^(2/3) (2^(1/2) +3^(1/3) )^4 =4+8×2^(1/2) ×3^(1/3) +12×3^(2/3) +12×2^(1/2) +3×3^(1/3) (2^(1/2) +3^(1/3) )^5 =60+4×2^(1/2) +20×3^(1/3) +20×2^(1/2) ×3^(2/3) +15×2^(1/2) ×3^(1/3) +3×3^(2/3) (2^(1/2) +3^(1/3) )^6 =17+24×2^(1/2) ×3^(1/3) +60×3^(2/3) +120×2^(1/2) +90×3^(1/3) +18×2^(1/2) ×3^(2/3) determinant (((n \),1,2^(1/2) ,3^(1/3) ,3^(2/3) ,(2^(1/2) ×3^(1/3) ),(2^(1/2) ×3^(2/3) ),),(((...)^0 ),1,,,,,,a_0 ),(((...)^1 ),,1,1,,,,a_1 ),(((...)^2 ),2,,,1,2,,a_2 ),(((...)^3 ),3,2,6,,,3,a_3 ),(((...)^4 ),4,(12),3,(12),8,,a_4 ),(((...)^5 ),(60),4,(20),3,(15),(20),a_5 ),(((...)^6 ),(17),(120),(90),(60),(24),(18),)) a_0 +2a_2 +3a_3 +4a_4 +60a_5 +17=0 a_1 +2a_3 +12a_4 +4a_5 +120=0 a_1 +6a_3 +3a_4 +20a_5 +90=0 a_2 +12a_4 +3a_5 +60=0 2a_2 +8a_4 +15a_5 +24=0 3a_3 +20a_5 +18=0 solving this equation system we get a_0 =1, a_1 =−36, a_2 =12, a_3 =−6, a_4 =−6, a_5 =0, a_6 =1 (given) sum of all coefficients Σ_(n=0) ^6 a_n =−34 P(x)=x^6 −6x^4 −6x^3 +12x^2 −36x+1
$${say}\:{P}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{\mathrm{6}} {x}^{\mathrm{6}} \\ $$$${with}\:{a}_{\mathrm{6}} =\mathrm{1} \\ $$$${P}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}{a}_{{n}} \left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{{n}} =\mathrm{0} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{1}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{2}} =\mathrm{2}+\mathrm{2}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} =\mathrm{3}+\mathrm{2}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{6}×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{4}} =\mathrm{4}+\mathrm{8}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{12}×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{12}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{5}} =\mathrm{60}+\mathrm{4}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{20}×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{20}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{15}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{6}} =\mathrm{17}+\mathrm{24}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{60}×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{120}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{90}×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{18}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{{n}\:\backslash}&\hline{\mathrm{1}}&\hline{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} }&\hline{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }&\hline{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} }&\hline{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }&\hline{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} }&\hline{}\\{\left(…\right)^{\mathrm{0}} }&\hline{\mathrm{1}}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{{a}_{\mathrm{0}} }\\{\left(…\right)^{\mathrm{1}} }&\hline{}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{}&\hline{}&\hline{}&\hline{{a}_{\mathrm{1}} }\\{\left(…\right)^{\mathrm{2}} }&\hline{\mathrm{2}}&\hline{}&\hline{}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{}&\hline{{a}_{\mathrm{2}} }\\{\left(…\right)^{\mathrm{3}} }&\hline{\mathrm{3}}&\hline{\mathrm{2}}&\hline{\mathrm{6}}&\hline{}&\hline{}&\hline{\mathrm{3}}&\hline{{a}_{\mathrm{3}} }\\{\left(…\right)^{\mathrm{4}} }&\hline{\mathrm{4}}&\hline{\mathrm{12}}&\hline{\mathrm{3}}&\hline{\mathrm{12}}&\hline{\mathrm{8}}&\hline{}&\hline{{a}_{\mathrm{4}} }\\{\left(…\right)^{\mathrm{5}} }&\hline{\mathrm{60}}&\hline{\mathrm{4}}&\hline{\mathrm{20}}&\hline{\mathrm{3}}&\hline{\mathrm{15}}&\hline{\mathrm{20}}&\hline{{a}_{\mathrm{5}} }\\{\left(…\right)^{\mathrm{6}} }&\hline{\mathrm{17}}&\hline{\mathrm{120}}&\hline{\mathrm{90}}&\hline{\mathrm{60}}&\hline{\mathrm{24}}&\hline{\mathrm{18}}&\hline{}\\\hline\end{array} \\ $$$${a}_{\mathrm{0}} +\mathrm{2}{a}_{\mathrm{2}} +\mathrm{3}{a}_{\mathrm{3}} +\mathrm{4}{a}_{\mathrm{4}} +\mathrm{60}{a}_{\mathrm{5}} +\mathrm{17}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} +\mathrm{2}{a}_{\mathrm{3}} +\mathrm{12}{a}_{\mathrm{4}} +\mathrm{4}{a}_{\mathrm{5}} +\mathrm{120}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} +\mathrm{6}{a}_{\mathrm{3}} +\mathrm{3}{a}_{\mathrm{4}} +\mathrm{20a}_{\mathrm{5}} +\mathrm{90}=\mathrm{0} \\ $$$${a}_{\mathrm{2}} +\mathrm{12}{a}_{\mathrm{4}} +\mathrm{3}{a}_{\mathrm{5}} +\mathrm{60}=\mathrm{0} \\ $$$$\mathrm{2}{a}_{\mathrm{2}} +\mathrm{8}{a}_{\mathrm{4}} +\mathrm{15}{a}_{\mathrm{5}} +\mathrm{24}=\mathrm{0} \\ $$$$\mathrm{3}{a}_{\mathrm{3}} +\mathrm{20}{a}_{\mathrm{5}} +\mathrm{18}=\mathrm{0} \\ $$$${solving}\:{this}\:{equation}\:{system}\:{we}\:{get} \\ $$$${a}_{\mathrm{0}} =\mathrm{1},\: \\ $$$${a}_{\mathrm{1}} =−\mathrm{36},\: \\ $$$${a}_{\mathrm{2}} =\mathrm{12},\: \\ $$$${a}_{\mathrm{3}} =−\mathrm{6},\: \\ $$$${a}_{\mathrm{4}} =−\mathrm{6},\: \\ $$$${a}_{\mathrm{5}} =\mathrm{0},\: \\ $$$${a}_{\mathrm{6}} =\mathrm{1}\:\left({given}\right) \\ $$$${sum}\:{of}\:{all}\:{coefficients}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}{a}_{{n}} =−\mathrm{34} \\ $$$${P}\left({x}\right)={x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{36}{x}+\mathrm{1} \\ $$

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