<s>[INST] If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$. Question from Math Olympiad:

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

My approach:
$a\sqrt{a} + b\sqrt {b} = 183$
$a\sqrt{b} + b\sqrt {a} = 182$
Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$
I only see two integral possibilities:
$1 \times 365 $  and $5 \times 73 $
And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.
Please help me with the general method to go about the same.

This is the picture from the source.
 [/INST] Let $\sqrt a=x,\sqrt b=y$
$$\implies\dfrac{183}{182}=\dfrac{x^3+y^3}{xy(x+y)}$$
As $x+y\ne0,$ $$182 x^2-365xy+182y^2=0\implies\dfrac xy=\dfrac{13}{14}\text{ or }\dfrac{14}{13}$$
If $\dfrac xy=\dfrac{13}{14},\dfrac x{13}=\dfrac y{14}=z$(say)
So, we have $$182=z^3(13\cdot14)(13+14)\implies z^3=?$$
As $z$ is real, $z=\sqrt[3]{\dfrac1{27}}=\dfrac13\implies x=?,y=?$
We need $$\dfrac{9(a+b)}5=\dfrac{9(x^2+y^2)}5=?$$
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