<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] Hint: Let $x=\sqrt a$ and $y=\sqrt b$. You know that
$$x^3+y^3=183$$
and
$$xy(x+y)=182.$$
Since
$$x^3+3xy(x+y)+y^3=(x+y)^3,$$
you can use this to find $x+y$; see if you can use this to find $\frac 95(x^2+y^2).$
</s>