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Question Number 119119 by benjo_mathlover last updated on 22/Oct/20

Answered by mindispower last updated on 22/Oct/20

⇒abc+1=(a/(a+b+c))...1 abc+2=(b/(a+b+c))....2 abc+7=(c/(a+b+c))...3 abc+10=1 ⇒abc=−9...2 ⇒bc=−(9/a),ac=((−9)/b) bc+(1/a)=−(8/a)=−(7/b)=−(2/c) b=((8a)/7),c=(a/4) 2⇔((8a^3 )/(28))=−9⇒a^3 =((−63)/4) a=−(((63)/4))^(1/3) ,real solution ,z=(√((63)/4))e^((2iπ)/3) ,z^−

⇒abc+1=aa+b+c...1abc+2=ba+b+c....2abc+7=ca+b+c...3abc+10=1⇒abc=−9...2⇒bc=−9a,ac=−9bbc+1a=−8a=−7b=−2cb=8a7,c=a42⇔8a328=−9⇒a3=−634a=−6343,realsolution,z=634e2iπ3,z−

Answered by bemath last updated on 22/Oct/20

(i) ((abc+1)/a) = (1/(a+b+c)) ⇒ abc+1 = (a/(a+b+c)) (ii) ((abc+2)/b) = (1/(a+b+c)) ⇒abc+2 = (b/(a+b+c)) (iii) ((abc+7)/c) = (1/(a+b+c))⇒ abc+7 = (c/(a+b+c)) ⇒(i)+(ii)+(iii) 3abc + 10 = 1 ⇒abc = −3 → { ((ab=−(3/c))),((ac=−(3/b) )),((bc=−(3/a))) :} ⇒ bc +(1/a) = (1/(a+b+c)) ; −(2/a) = (1/(a+b+c)) ⇒ac+(2/b) = (1/(a+b+c)) ; −(1/b)=(1/(a+b+c)) ⇒ab+(7/c)= (1/(a+b+c)) ; (4/c) = (1/(a+b+c)) ⇔ −(2/a) = −(1/b) = (4/c) let a = k → { ((b= (k/2))),((c=−2k)) :} ⇒ ((−k^3 +1)/k) = (1/(−(k/2))) ; k^3 = 3 , k = (3)^(1/(3 )) therefore { ((a=(3)^(1/(3 )) )),((b=((3)^(1/(3 )) /2) )),((c=−2 (3)^(1/(3 )) )) :}

(i)abc+1a=1a+b+c⇒abc+1=aa+b+c(ii)abc+2b=1a+b+c⇒abc+2=ba+b+c(iii)abc+7c=1a+b+c⇒abc+7=ca+b+c⇒(i)+(ii)+(iii)3abc+10=1⇒abc=−3→{ab=−3cac=−3bbc=−3a⇒bc+1a=1a+b+c;−2a=1a+b+c⇒ac+2b=1a+b+c;−1b=1a+b+c⇒ab+7c=1a+b+c;4c=1a+b+c⇔−2a=−1b=4cleta=k→{b=k2c=−2k⇒−k3+1k=1−k2;k3=3,k=33therefore{a=33b=332c=−233

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