Algebra> If a/(b+c)+b/(c+a)+c/(a+b)=1,then a2/(b+c...

If a/(b+c)+b/(c+a)+c/(a+b)=1,then a2/(b+c)+b2/(c+a)+c2/(a+b)=?

Tushar kanti Das , 9 Years ago

Grade 10

2 Answers

jagdish singh singh

$\hspace{-.7 cm}$Given $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} = 1\;,$ Multiply both side by $a+b+c$\\\\\\ So $\bigg(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg)\cdot (a+b+c)=a+b+c$\\\\\\ $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{a(b+c)}{b+c}+\frac{b(c+a)}{c+a}+\frac{c(a+b)}{a+b}=a+b+c$\\\\\\ So we get $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} = 0$$

Last Activity: 9 Years ago

Rishi Sharma

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

Last Activity: 5 Years ago