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