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Question Number 36019 by math1967 last updated on 27/May/18
If a+b+c=0 show that ((a/(b−c))+(b/(c−a))+(c/(a−b)))(((b−c)/a)+((c−a)/b) +((a−b)/c))=9
Ifa+b+c=0showthat(ab−c+bc−a+ca−b)(b−ca+c−ab+a−bc)=9
Answered by ajfour last updated on 27/May/18
let b=pa , c=qa ⇒ a(1+p+q)=0 ...(i) (shall be using this in many steps that follow..) l.h.s. = ((1/(p−q))+(p/(q−1))+(q/(1−p)))(((p−q)/1)+((q−1)/p)+((1−p)/q)) =1+(((q−1))/(p(p−q)))+(((1−p))/(q(p−q)))+ ((p(p−q))/(q−1))+1+((p(1−p))/(q(q−1)))+ ((q(p−q))/((1−p)))+((q(q−1))/(p(1−p)))+1 =3+((q^2 −q+p−p^2 )/(pq(p−q)))+((pq(p−q)+p(1−p))/(q(q−1)))+ ((pq(p−q)+q(q−1))/(p(1−p))) =3+((1−p−q)/(pq))+((p[qp−q^2 +1−p])/(q(q−1)))+ ((q[p^2 −pq+q−1])/(p(1−p))) =3+(2/(pq))+((p(p−q−1))/q)+((q(q−p−1))/p) =3+((2+p^2 (p−q−1)+q^2 (q−p−1))/(pq)) =3+((2+p^3 +q^3 −pq(p+q)−(p^2 +q^2 ))/(pq)) =3+((2−(p^2 +q^2 −pq)+pq−1+2pq)/(pq)) =3+((2−1+2pq+pq+pq−1+2pq)/(pq)) = 3+6 = 9 .
letb=pa,c=qa⇒a(1+p+q)=0…(i)(shallbeusingthisinmanystepsthatfollow..)l.h.s.=(1p−q+pq−1+q1−p)(p−q1+q−1p+1−pq)=1+(q−1)p(p−q)+(1−p)q(p−q)+p(p−q)q−1+1+p(1−p)q(q−1)+q(p−q)(1−p)+q(q−1)p(1−p)+1=3+q2−q+p−p2pq(p−q)+pq(p−q)+p(1−p)q(q−1)+pq(p−q)+q(q−1)p(1−p)=3+1−p−qpq+p[qp−q2+1−p]q(q−1)+q[p2−pq+q−1]p(1−p)=3+2pq+p(p−q−1)q+q(q−p−1)p=3+2+p2(p−q−1)+q2(q−p−1)pq=3+2+p3+q3−pq(p+q)−(p2+q2)pq=3+2−(p2+q2−pq)+pq−1+2pqpq=3+2−1+2pq+pq+pq−1+2pqpq=3+6=9.
Commented by tanmay.chaudhury50@gmail.com last updated on 27/May/18
praisworthy...
praisworthy…
Answered by ajfour last updated on 27/May/18
l.h.s. = 3+Σ((a/(b−c)))[(((c−a)/b))+(((a−b)/c))] =3+Σ((a/(b−c)))(((c^2 −ac+ab−b^2 )/(bc))) =3+Σ((a^2 /(abc)))(a−b−c) =3+Σ((a^2 /(abc)))(2a) ∵ (−b−c=a) =3+2Σ(a^3 /(abc)) and as Σa^3 =3abc if Σa=0, so l.h.s. =3+2(((3abc)/(abc)))= 9 .
l.h.s.=3+Σ(ab−c)[(c−ab)+(a−bc)]=3+Σ(ab−c)(c2−ac+ab−b2bc)=3+Σ(a2abc)(a−b−c)=3+Σ(a2abc)(2a)∵(−b−c=a)=3+2Σa3abcandasΣa3=3abcifΣa=0,sol.h.s.=3+2(3abcabc)=9.
Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
zLHS 3+(a/(b−c))×(((c−a)/b)+((a−b)/c))+(b/(c−a))×(((b−c)/a)+((a−b)/c))+ (c/(a−b))×(((b−c)/a)+((c−a)/b)) let us calculate (a/(b−c))×(((c−a)/b)+((a−b)/c)) (a/(b−c))×(((c^2 −ac+ab−b^2 )/(bc))) =(a/(b−c))×{((a(b−c)−(b^2 −c^2 ))/(bc))} =(a/(b−c))×{(((b−c)(a−b−c))/(bc))} =(a/(bc))×(a−b−c) =(a/(bc))×2a since a+b+c=0 so (−b−c)=a =2(a^3 /(abc)) other two also give ((2b^3 )/(abc)) and ((2c^3 )/(abc)) so 3+((2a^3 )/(abc))+((2b^3 )/(abc))+((2c^3 )/(abc)) 3+2(((a^3 +b^3 +c^3 )/(abc))) 3+2(((3abc)/(abc))) =3+6 =9 if a+b+c=0 a^3 +b^3 +c^3 =3abc
zLHS3+ab−c×(c−ab+a−bc)+bc−a×(b−ca+a−bc)+ca−b×(b−ca+c−ab)letuscalculateab−c×(c−ab+a−bc)ab−c×(c2−ac+ab−b2bc)=ab−c×{a(b−c)−(b2−c2)bc}=ab−c×{(b−c)(a−b−c)bc}=abc×(a−b−c)=abc×2asincea+b+c=0so(−b−c)=a=2a3abcothertwoalsogive2b3abcand2c3abcso3+2a3abc+2b3abc+2c3abc3+2(a3+b3+c3abc)3+2(3abcabc)=3+6=9ifa+b+c=0a3+b3+c3=3abc

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