Positive real numbers and their equations

Given : $ \displaystyle a,b$ are positive real numbers such that $ \displaystyle a\sqrt{a}+b\sqrt{b}=183$ and $ \displaystyle a\sqrt{b}+b\sqrt{a}=182$ .
Find : $ \displaystyle \frac{9}{5}\ (\ a+b\ )$

Solution

$ \displaystyle a\sqrt{a}+b\sqrt{b}=183$

$ \displaystyle {{(\ \sqrt{a}\ )}^{3}}+{{(\ \sqrt{b}\ )}^{3}}=183$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-3\sqrt{{ab}}(\ \sqrt{a}+\sqrt{b}\ )=183\ ........(1)$

$ \displaystyle a\sqrt{b}+b\sqrt{a}=182$ 

$ \displaystyle \sqrt{{ab}}\ (\ \sqrt{a}+\sqrt{b}\ )=182\ \ .............(2)$ 

$ \displaystyle \text{From}\ (1)\ \text{and}\ (2),$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-546=183$

$ \displaystyle \ \ \ \ \ \ \ \ {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}=729$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{a}+\sqrt{b}=9\ \ \ .............(3)$

$ \displaystyle \text{From}\ (2)\ \text{and}\ (3),$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \sqrt{{ab}}=\frac{{182}}{9}$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{2}}=81\ \ \ \ \left[ {By\ (3)} \right]$

$ \displaystyle a+b+2\sqrt{{ab}}=81$

$ \displaystyle \ \ a+b+\frac{{364}}{9}=81$

$ \displaystyle \ a+b=\frac{{365}}{9}=73\times \frac{5}{9}$

$ \displaystyle \therefore \ \ \frac{9}{5}\ (\ a+b\ )=73$