Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$

Question

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

It is my equation. I subtracted the second equation from the first one and found $(a-b)(\sqrt{a}-\sqrt{b})=1$.or, $(a-b)^2=(\sqrt{a}+\sqrt{b})$.

Am I going to the right path? How to escape from this. Please give me hints. Thank you.

Answer 1

Hint:  let $\,u=\sqrt{a}\,$ and $\,v=\sqrt{b}\,$, then it's given that $\,u^3+v^3=183\,$ and $\,u^2v+uv^2=182\,$, and the problem asks for $\frac{9}{5}(u^2+v^2)$.

• $\;(u+v)^3=u^3 + v^3 + 3(u^2v+ uv^2) = 183 + 3 \cdot 182 = 729 \implies u+v = 9\,$

• $\;uv= (u^2v+uv^2)/(u+v) = 182 / 9 \,$

• $\;u^2+v^2 = (u+v)^2 - 2 uv = \cdots \,$

Answer 2

Let $A^2=a$ and $B^2=b$ then we have \begin{eqnarray*} A^3+B^3 = 183 \\ A^2B+B^2 A =182 \end{eqnarray*} Multiply the second equation by $3$ and add the first \begin{eqnarray*} A^3+B^3+ 3(A^2B+B^2 A) =(A+B)^3= 729 \\ A+B =9 \\ AB = \frac{182}{9} \\ (A-B)^2=(A+B)^2-4AB= \frac{1}{9} \\ A= \frac{14}{3} \\ B= \frac{13}{3} \end{eqnarray*} So $ \frac{9}{5} (a+b) = \color{red}{73}$.