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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)

(122+133)=(14+127)=(27108+4108)therootis31108sumofcoeffecients=P(1)letQ(n)=108P(n)P(1)=Q(1)108letQ(n)=a6n6+a5n5...+a1n+a0P(1)=Q(1)/108didyouprovideaspecificpolynomial?ifnotthesumdependsonQ(n)

Commented by Abdullahrussell last updated on 08/Jan/25

 2^(1/2) +3^(1/3) =(√2) +(3)^(1/3)

212+313=2+33

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

sayP(x)=a0+a1x+a2x2+...+a6x6witha6=1P(212+313)=6n=0an(212+313)n=0(212+313)1=212+313(212+313)2=2+2×212×313+323(212+313)3=3+2×212+6×313+3×212×323(212+313)4=4+8×212×313+12×323+12×212+3×313(212+313)5=60+4×212+20×313+20×212×323+15×212×313+3×323(212+313)6=17+24×212×313+60×323+120×212+90×313+18×212×323n1212313323212×313212×323(...)01a0(...)111a1(...)2212a2(...)33263a3(...)44123128a4(...)56042031520a5(...)61712090602418a0+2a2+3a3+4a4+60a5+17=0a1+2a3+12a4+4a5+120=0a1+6a3+3a4+20a5+90=0a2+12a4+3a5+60=02a2+8a4+15a5+24=03a3+20a5+18=0solvingthisequationsystemwegeta0=1,a1=36,a2=12,a3=6,a4=6,a5=0,a6=1(given)sumofallcoefficients6n=0an=34P(x)=x66x46x3+12x236x+1

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