Algebra> if a/b+c+b/c+a+c/a+b=1 then find a*a/b+c+...

if a/b+c+b/c+a+c/a+b=1 then find aa/b+c+bb/c+a+c*c/a+b

jairam singh , 7 Years ago

Grade 12th pass

4 Answers

Arun

Last Activity: 7 Years ago

Given relation

a/(b+c)+b/(c+a)+c/(a+b)=1

Multiplying both sides by a+b+c we get (a+b+c!=0)

⇒a(a+b+c)b+c+b(a+b+c)c+a+c(a+b+c)a+b=(a+b+c)=>(a(a+b+c))/(b+c)+(b(a+b+c))/(c+a)+(c(a+b+c))/(a+b)=(a+b+c)

⇒a2b+c+a(b+c)b+c+b2c+a+b(c+a)c+a+c2a+b+c(a+b)a+b=(a+b+c)=>a^2/(b+c)+(a(b+c))/(b+c)+b^2/(c+a)+(b(c+a))/(c+a)+c^2/(a+b)+(c(a+b))/(a+b)=(a+b+c)

⇒a2b+c+a+b2c+a+b+c2a+b+c=(a+b+c)=>a^2/(b+c)+a+b^2/(c+a)+b+c^2/(a+b)+c=(a+b+c)

=>a^2/(b+c)+b^2/(c+a) + c^2/(a+b) = 0

Regards

Arun (askIITians forum expert)

sohel mia

Last Activity: 6 Years ago

We have multiply both sides of it. a(a+b+c) /(b+c) +b(a+b +c) /(c+a) +c(a+b+c) /(a+b) =(a+b+c). Or, [{a²+a(b+c)} ÷(b+c)] +[{b²+b(a+c)} ÷(c+a)] +[{c²+c(a+b)}] =(a+b+c) Or {a²÷

sohel mia

Last Activity: 6 Years ago

We have multiply both sides of it by (a+b+c). {a(a+b+c) /(b+c)} +{b(a+b +c)/(a+c)} +{c(a+b+c) /(a+b)}=(a+b+c) Or, [{a²+a(b+c)}/(b+c)]+[{b²+b(a+c)}/(a+c)]+[{c²+c(a+b)}/(a+b)] =(a+b+c) Or, {a²/(b+c)}+a+{b²/(c+a)}+b+{c²/(a+b)}+c=a+b+c Or, [{a²/(b+c)}+{b²/(c+a)}+{c²/(a+b)}]+(a+b+c)=(a+b+c). At least, [{a²/(b+c)}+{b²/(c+a)}+{c²/(a+b)}] =0 (proved)

Rishi Sharma

Last Activity: 5 Years ago

Dear Student,
Please find below the solution to your problem.

We know that
a/(b+c) + b/(c+a) + c/(a+b) −1 = 0 which is equivalent to
a^3 + b^3 + abc + c^3 / (a+b)(a+c)(b+c) = 0
also
= a^2/(b+c) + b^2/(c+a) + c^2/(a+b)
= a^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4 /(a+b)(a+c)(b+c) = 0
but
a^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4 = (a+b+c)(a^3 + b^3 + abc + c^3)
so
a^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4=0
and consequently a^2/(b+c) + b^2/(c+a) + c^2/(a+b) = 0

Thanks and Regards