Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$ 
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$, $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.

It is my equation. I subtracted the second equation from the first one and found $(a-b)(\sqrt{a}-\sqrt{b})=1$.or, $(a-b)^2=(\sqrt{a}+\sqrt{b})$.      
Am I going to the right path? How to escape from this. Please give me hints. Thank you.

Hint:  let $\,u=\sqrt{a}\,$ and $\,v=\sqrt{b}\,$, then it's given that $\,u^3+v^3=183\,$ and $\,u^2v+uv^2=182\,$, and the problem asks for $\frac{9}{5}(u^2+v^2)$.


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*$\;(u+v)^3=u^3 + v^3 + 3(u^2v+ uv^2) = 183 + 3 \cdot 182 = 729 \implies u+v = 9\,$

*$\;uv= (u^2v+uv^2)/(u+v) = 182 / 9  \,$

*$\;u^2+v^2 = (u+v)^2 - 2 uv = \cdots \,$
