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

This problem will use the following elementary algebraic identity: $$ (x+y)^3 = x^3 + y^3 + 3x^2y + 3xy^2 $$

Can you identify what is x and what is y?

Set \( x = \sqrt a, y = \sqrt b \). Then the given information translates to $$ x^3 + y^3 = 183 , x^2y + xy^2 = 182 $$

This implies

\( (x + y)^3 \\ = ( \sqrt a + \sqrt b)^3 \\ = x^3 + y^3 + 3(x^2y + xy^2) \\ = 183 + 3 \times 182 \\= 729 \)

Finally taking cube root on both sides, we have \( \sqrt a + \sqrt b = 9 \)

Note that

\( \sqrt a b + a \sqrt b = 182 \\ \Rightarrow \sqrt a \sqrt b ( \sqrt a + \sqrt b ) = 182 \\ \Rightarrow \sqrt {ab} \times 9 = 182 \)

So at this point we know \( ( \sqrt a + \sqrt b ) = 9, \sqrt{ab} = \frac{182}{9} \). It should be easy to find the value of \( \frac{9}(5) (a+b) \) from these relations.

 

\( a+b \\ = (\sqrt a + \sqrt b )^2 – 2 \sqrt{ab} \\ = 9^2 – 2 \times \frac {182}{9} \\ = \frac{365}{9} \)

Hence \( \frac {9}{5}(a+b) = \frac {9}{5} \times \frac {365}{9} = 73 \)