Problem
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)).
Hint 1
This problem will use the following elementary algebraic identity:
$(x+y)^3=x^3+y^3+3 x^2 y+3 x y^2$
Can you identify what is x and what is y?
Hint 2
background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none"] Set $x=\sqrt{a}, y=\sqrt{b}$. Then the given information translates to
$$
x^3+y^3=183, x^2 y+x y^2=182
$$
This implies $(x+y)^3=(\sqrt{a}+\sqrt{b})^3=x^3+y^3+3\left(x^2 y+x y^2\right)=183+3 \times 182=729$ Finally taking cube root on both sides, we have $\sqrt{a}+\sqrt{b}=9$
Hint 3
Note that $\sqrt{a} b+a \sqrt{b}=182 \Rightarrow \sqrt{a} \sqrt{b}(\sqrt{a}+\sqrt{b})=182 \Rightarrow \sqrt{a b} \times 9=182$ So at this point we know $(\sqrt{a}+\sqrt{b})=9, \sqrt{a b}=\frac{182}{9}$. It should be easy to find the value of $\left.\frac{9}{( } 5\right)(a+b)$ from these relations.
Final Answer
$a+b=(\sqrt{a}+\sqrt{b})^2-2 \sqrt{a b}=9^2-2 \times \frac{182}{9}=\frac{365}{9}$
Hence $\frac{9}{5}(a+b)=\frac{9}{5} \times \frac{365}{9}=73$
