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Question Number 50163 by Meritguide1234 last updated on 14/Dec/18

Answered by peter frank last updated on 14/Dec/18

let x=((b−c)/a) (1/x)=(a/(b−c)) y=((c−a)/b) (1/y)=(b/(c−a)) z=((a−b)/c) (1/z)=(c/(a−b)) (x+y+z)((1/x)+(1/y)+(1/z)) 3+(x/y)+(y/x)+(z/x)+(x/z)+(y/z)+(z/y) pls wait...

letx=b−ca1x=ab−cy=c−ab1y=bc−az=a−bc1z=ca−b(x+y+z)(1x+1y+1z)3+xy+yx+zx+xz+yz+zyplswait...

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

(p+q+r)((1/p)+(1/q)+(1/r)) =3+p((1/q)+(1/r))+q((1/r)+(1/p))+r((1/p)+(1/q)) =3+(p/q)+(r/q)+(p/r)+(q/r)+(q/p)+(r/p) =3+((p+r)/q)+((p+q)/r)+((q+r)/p) ((p+r)/q)=((((b−c)/a)+((a−b)/c))/((c−a)/b)) =((b(bc−c^2 +a^2 −ab))/(ac(c−a))) =(b/(ac))×((b(c−a)−(c+a)(c−a))/((c−a))) =(b/(ac))×(((c−a)(b−c−a))/((c−a))) =(b/(ac))×2b =((2b^3 )/(abc)) similarly others are ((2c^3 )/(abc)),((2a^3 )/(abc)) so 3+((2b^3 )/(abc))+((2c^3 )/(abc))+((2a^3 )/(abc)) =3+(2/(abc))(a^3 +b^3 +c^3 ) =3+(2/(abc))×3abc [when a+b+c=0 , a^3 +b^3 +c^3 =3abc] =3+6=9

(p+q+r)(1p+1q+1r)=3+p(1q+1r)+q(1r+1p)+r(1p+1q)=3+pq+rq+pr+qr+qp+rp=3+p+rq+p+qr+q+rpp+rq=b−ca+a−bcc−ab=b(bc−c2+a2−ab)ac(c−a)=bac×b(c−a)−(c+a)(c−a)(c−a)=bac×(c−a)(b−c−a)(c−a)=bac×2b=2b3abcsimilarlyothersare2c3abc,2a3abcso3+2b3abc+2c3abc+2a3abc=3+2abc(a3+b3+c3)=3+2abc×3abc[whena+b+c=0,a3+b3+c3=3abc]=3+6=9

Commented by peter frank last updated on 14/Dec/18

nice work sir

niceworksir

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

thank you...

thankyou...

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