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

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.

Answer 1

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 $\frac 95(x^2+y^2).$

Answer 2

Hint : If you make the difference of the two equations, you get $$(\sqrt{a}-\sqrt{b})(a-b)=0$$

Probably you can conclude.

Answer 3

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 $$\dfrac{9(a+b)}5=\dfrac{9(x^2+y^2)}5=?$$