Question

Please solve this problem​

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given

         $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$

• Multiply both sides by $a+b+c$

     $\Rightarrow \dfrac{a(a+b+c)}{b+c}+\dfrac{b(a+b+c)}{c+a}+\dfrac{c(a+b+c)}{a+b}=(a+b+c)\\\\\Rightarrow \dfrac{a^2+a(b+c))}{b+c}+\dfrac{b^2+b(c+a)}{c+a}+\dfrac{c^2+c(a+b)}{a+b}=(a+b+c)$

• Split the terms

   $\Rightarrow \dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=(a+b+c)$

• Cancel $a+b+c$ from both sides

   $\Rightarrow \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$

Hence proved