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 $\frac{l}{(a-c)(c-\sqrt{ab})}=\frac{m}{(b-c)(a-\sqrt{bc})}=\frac{n}{(c-a)(b-\sqrt{ac})}.$

Answer 1

Given,

$\frac{l}{\sqrt{a}-\sqrt{b}}+\frac{m}{\sqrt{b}-\sqrt{c}}+\frac{n}{\sqrt{c}-\sqrt{a}}=0,\frac{l}{\sqrt{a}+\sqrt{b}}+\frac{m}{\sqrt{b}+\sqrt{c}}+\frac{n}{\sqrt{c}+\sqrt{a}}=0$.

By cross multiplication we have,

$\frac{l}{A}=\frac{m}{B}=\frac{n}{C}$, where

$A=\frac{1}{\sqrt{b}-\sqrt{c}}\times \frac{1}{\sqrt{c}+\sqrt{a}}-\frac{1}{\sqrt{b}+\sqrt{c}}\times \frac{1}{\sqrt{c}-\sqrt{a}}=\frac{c-\sqrt{ab}}{(b-c)(c-a)}=\frac{(a-b)(c-\sqrt{ab})}{(a-b)(b-c)(c-a)}$.

Similarly, $B=\frac{(b-c)(a-\sqrt{bc})}{(a-b)(b-c)(c-a)}$ and $C=\frac{(c-a)(b-\sqrt{ac})}{(a-b)(b-c)(c-a)}$.

Therefore $\frac{l}{\frac{(a-b)(c-\sqrt{ab})}{(a-b)(b-c)(c-a)}}=\frac{m}{\frac{(b-c)(a-\sqrt{bc})}{(a-b)(b-c)(c-a)}}=\frac{n}{\frac{(c-a)(b-\sqrt{ac})}{(a-b)(b-c)(c-a)}}$

Or $\frac{l}{(a-b)(c-\sqrt{ab})}=\frac{m}{(b-c)(a-\sqrt{bc})}=\frac{n}{(c-a)(b-\sqrt{ac})}$.

Therefore, hence showed.