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.

Let $A^2=a$ and $B^2=b$ then we have
\begin{eqnarray*}
A^3+B^3 = 183 \\
A^2B+B^2 A =182
\end{eqnarray*}
Multiply the second equation by $3$ and add the first
\begin{eqnarray*}
A^3+B^3+ 3(A^2B+B^2 A)  =(A+B)^3= 729 \\
A+B =9 \\
AB = \frac{182}{9} \\
(A-B)^2=(A+B)^2-4AB= \frac{1}{9} \\
A= \frac{14}{3} \\
B= \frac{13}{3} 
\end{eqnarray*}
So $ \frac{9}{5} (a+b) = \color{red}{73}$.
