properties of ratio and rule of cross multiplication

Question

If $$\frac{l}{\sqrt a-\sqrt b}+\frac{m}{\sqrt b-\sqrt c}+\frac{n}{\sqrt c-\sqrt a} =0$$ $$\frac{l}{\sqrt a+\sqrt b}+\frac{m}{\sqrt b+\sqrt c}+\frac{n}{\sqrt c+\sqrt a} =0$$ Show that $$ \frac{l}{(a-b)(c-\sqrt ab)}=\frac{m}{(b-c)(a-\sqrt bc)}=\frac{n}{(c-a)(b-\sqrt ac)}$$ I try to solve it with all the properties of ratio and rule of cross muliplication ,which I know. I get $(a-b)$ term in denominator but I can't able to get $(c-\sqrt ab)$ term. I want to solve it on my own so give me hint to solve it. Thanks in advance.

Answer 1

For $a,b,c>0,$ let $a =A^2$ etc.

As $(B+C)(C-A)-(B-C)(C+A)=2(C^2-AB),$

$$\dfrac l{\dfrac n{(B-C)(C+A)}-\dfrac n{(B+C)(C-A)}}=\dfrac m{\cdots}=\dfrac1{\dfrac1{(A-B)(B+C)}-\dfrac1{(A+B)(B-C)}}$$

$$\implies\dfrac {l(B^2-C^2)(C^2-A^2)}{2(C^2-AB)}=\dfrac m{\cdots}=\dfrac{n(A^2-B^2)(B^2-C^2)}{2(B^2-CA)}$$

$$\iff\dfrac {l(C^2-A^2)}{C^2-AB}=\dfrac{n(A^2-B^2)}{B^2-CA}$$

Can you take it from here?