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# Real Surds – Problem 2 Pre RMO 2017

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 $\frac {9}{5}(a+b) = \frac {9}{5} \times \frac {365}{9} = 73$ ## By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

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