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a-1-3-b-1-3-c-1-3-1-3-1-3-2-1-3-b-c-a-




Question Number 217643 by Rojarani last updated on 17/Mar/25
 a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) ))    b+c−a=?
a13+b13+c13=1313213b+ca=?
Answered by mehdee7396 last updated on 17/Mar/25
(1/( (3)^(1/3) −(2)^(1/3) ))=(9)^(1/3) +(6)^(1/3) +(4)^(1/3) ⇒  a=9 &b=6&c=4⇒b+c−a=1  or  a=6 &b=9&c=4u⇒b+c−a=7  or  a=4 &b=9&c=6⇒b+c−a=11  1 or  7  or  11
13323=93+63+43a=9&b=6&c=4b+ca=1ora=6&b=9&c=4ub+ca=7ora=4&b=9&c=6b+ca=111or7or11
Commented by Rojarani last updated on 17/Mar/25
Sir thanks.
Sirthanks.
Answered by Rasheed.Sindhi last updated on 17/Mar/25
 a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) ))    b+c−a=?   a^(1/3) +b^(1/3) +c^(1/3)   =(1/(3^(1/3) −2^(1/3) )) ∙ ((3^(2/3) +3^(1/3) ∙2^(1/3) +2^(2/3) )/(3^(2/3) +3^(1/3) ∙2^(1/3) +2^(2/3) ))  =((9^(1/3) +6^(1/3) +4^(1/3) )/((3^(1/3) )^3 −(2^(1/3) )^3 ))=((9^(1/3) +6^(1/3) +4^(1/3) )/(3−2))             =9^(1/3) +6^(1/3) +4^(1/3)   Case a=9  b+c−a=6+4−9=1 ✓  Case a=6  b+c−a=9+4−6=7 ✓  Case a=4  b+c−a=9+6−4=11 ✓   determinant ((((A−B)(A^2 +AB+B^2 )=A^3 −B^3 ^(Formula used) )))
a13+b13+c13=1313213b+ca=?a13+b13+c13=1313213323+313213+223323+313213+223=913+613+413(313)3(213)3=913+613+41332=913+613+413Casea=9b+ca=6+49=1Casea=6b+ca=9+46=7Casea=4b+ca=9+64=11(AB)(A2+AB+B2)=A3B3Formulaused

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