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Question Number 130036 by liberty last updated on 22/Jan/21

If { ((a+b+c = 1)),((a^3 +b^3 +c^3 = 4)) :} find (1/(a+bc)) + (1/(b+ca)) + (1/(c+ab)).

If{a+b+c=1a3+b3+c3=4find1a+bc+1b+ca+1c+ab.

Answered by mnjuly1970 last updated on 22/Jan/21

solution: {_(a^3 +b^3 +c^3 =4) ^(a+b+c=1) K=(1/(a+bc))+(1/(b+ac)) +(1/(c+ab))=? a+b+c−1=0 a^3 +b^3 +(c−1)^3 =3ab(c−1) (euler identity) a^3 +b^3 +c^3 −3c^2 +3c−1=3abc−3ab 1−c^2 +c=ab(c−1) (cyclic) 1=(c−1)(c+ab)....(1) ∴ 1=(a−1)(a+bc)....(2) 1=(b−1)(b+ac)...(3) ∴ K=(1/(a+bc))+(1/(b+ca)) +(1/(c+ab)) K=^( (1),(2),(3)) a−1+b−1+c−1 K=a+b+c−3=1−3=−2 ✓

solution:{a3+b3+c3=4a+b+c=1K=1a+bc+1b+ac+1c+ab=?a+b+c−1=0a3+b3+(c−1)3=3ab(c−1)(euleridentity)a3+b3+c3−3c2+3c−1=3abc−3ab1−c2+c=ab(c−1)(cyclic)1=(c−1)(c+ab)....(1)∴1=(a−1)(a+bc)....(2)1=(b−1)(b+ac)...(3)∴K=1a+bc+1b+ca+1c+abK=(1),(2),(3)a−1+b−1+c−1K=a+b+c−3=1−3=−2✓

Answered by EDWIN88 last updated on 22/Jan/21

(1) a+b =1−c⇒(a+b)^3 =(1−c)^3 c^2 −c−ab(a+b)=1 (2)(a+b)^2 =(1−c)^2 ⇒a^2 +b^2 +c^2 =1−2c−2ab (a+b+c)^2 −2ab−2ac−2bc=1−2c−2ab c−ac−bc = 0 →c(1−(a+b))=0 a+b = 1 ∧ c = 0 (3) from (1/(a+bc)) + (1/(b+ca)) + (1/(c+ab)) = (1/a)+(1/b)+(1/(ab)) ((a+b)/(ab)) + (1/(ab)) = ((a+b+1)/(ab)) = (2/(ab)) and from eq(1) c^2 −c−ab(a+b)=1 give −ab = (1/(a+b)) ⇒ab=−(1/(a+b)) =−1 therefore we get (2/(ab)) = (2/(−1))=−2

(1)a+b=1−c⇒(a+b)3=(1−c)3c2−c−ab(a+b)=1(2)(a+b)2=(1−c)2⇒a2+b2+c2=1−2c−2ab(a+b+c)2−2ab−2ac−2bc=1−2c−2abc−ac−bc=0→c(1−(a+b))=0a+b=1∧c=0(3)from1a+bc+1b+ca+1c+ab=1a+1b+1aba+bab+1ab=a+b+1ab=2abandfromeq(1)c2−c−ab(a+b)=1give−ab=1a+b⇒ab=−1a+b=−1thereforeweget2ab=2−1=−2

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