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Grade: 12th pass
        if a/b+c+b/c+a+c/a+b=1 then find a*a/b+c+b*b/c+a+c*c/a+b
one year ago

Answers : (3)

Arun
21467 Points
							

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)

one year ago
sohel mia
16 Points
							
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÷
7 months ago
sohel mia
16 Points
							
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)
 
7 months ago
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