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Question Number 174933 by AgniMath last updated on 14/Aug/22

Answered by Rasheed.Sindhi last updated on 15/Aug/22

A=(((s−a)^2 )/((s−b)(s−c)))+(((s−b)^2 )/((s−a)(s−c)))+(((s−c)^2 )/((s−a)(s−b))) 2s=a+b+c, t=ab+bc+ca (say) A=(((s−a)^3 +(s−b)^3 +(s−c)^3 )/((s−a)(s−b)(s−c)))−3+3 =(((s−a)^3 +(s−b)^3 +(s−c)^3 −3(s−a)(s−b)(s−c)_(N) )/((s−a)(s−b)(s−c)_(D) ))+3 A=(N/D)+3 N=( (s−a)+(s−b)+(s−c)_(I) )((s−a)^2 +(s−b)^2 +(s−c)^2 _(II) −((s−a)(s−b)+(s−b)(s−c)+(s−a)(s−c)_(III) ) I=3s−(a+b+c)=3s−2s=s II=3s^2 +a^2 +b^2 +c^2 −2s(a+b+c) =3s^2 +(a+b+c)^2 −2(ab+bc+ca)−2s(2s) =3s^2 +(2s)^2 −2(t)−4s^2 =3s^2 −2t III=3s^2 +ab+bc+ca−2s(a+b+c) =3s^2 +t−4s^2 =−s^2 +t N=(s)( 3s^2 −2t−(−s^2 +t) ) =s(4s^2 −3t)=4s^3 −3st D=s^3 −s^2 (a+b+c)+s(ab+bc+ca)−abc =s^3 −s^2 (2s)+s(t)−abc =−s^3 +st−abc A=(N/D)+3⇒A=((4s^3 −3st)/(−s^3 +st−abc))+3 A=((4s^3 −3st−3s^3 +3st−3abc)/(−s^3 +st−abc)) =((s^3 −3abc)/(−s^3 +st−abc)) =(((((a+b+c)/2))^3 −3abc)/(−(((a+b+c)/2))^3 +(((a+b+c)/2))(ab+bc+ca)−abc)) A=(((a+b+c)^3 −24abc)/(−(a+b+c)^3 +4(a+b+c)(ab+bc+ca)−8abc))

A=(s−a)2(s−b)(s−c)+(s−b)2(s−a)(s−c)+(s−c)2(s−a)(s−b)2s=a+b+c,t=ab+bc+ca(say)A=(s−a)3+(s−b)3+(s−c)3(s−a)(s−b)(s−c)−3+3=(s−a)3+(s−b)3+(s−c)3−3(s−a)(s−b)(s−c)⏟N(s−a)(s−b)(s−c)⏟D+3A=ND+3Missing \left or extra \rightI=3s−(a+b+c)=3s−2s=sII=3s2+a2+b2+c2−2s(a+b+c)=3s2+(a+b+c)2−2(ab+bc+ca)−2s(2s)=3s2+(2s)2−2(t)−4s2=3s2−2tIII=3s2+ab+bc+ca−2s(a+b+c)=3s2+t−4s2=−s2+tN=(s)(3s2−2t−(−s2+t))=s(4s2−3t)=4s3−3stD=s3−s2(a+b+c)+s(ab+bc+ca)−abc=s3−s2(2s)+s(t)−abc=−s3+st−abcA=ND+3⇒A=4s3−3st−s3+st−abc+3A=4s3−3st−3s3+3st−3abc−s3+st−abc=s3−3abc−s3+st−abc=(a+b+c2)3−3abc−(a+b+c2)3+(a+b+c2)(ab+bc+ca)−abcA=(a+b+c)3−24abc−(a+b+c)3+4(a+b+c)(ab+bc+ca)−8abc

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