Let $\alpha,\beta$ and $\gamma$ be the positive roots of the cubic equation $x^3-ax^2+bx-c = 0$ such that $\alpha + 2\beta + 3\gamma =4$ and $81\alpha \beta\gamma = 32$ .

Given that the value of $b$ can be expressed as $\dfrac pq$ , where $p$ and $q$ are coprime positive integers , find $5q - 2p$ .

If you think there is insufficient information to solve this question, submit 185 as your answer.

The answer is 13.

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By $AM\geq GM$ . $\alpha+2\beta+3\gamma=4\geq 3\sqrt[3]{\alpha\cdot2\beta\cdot3\gamma}$ Some simplification gives:- $\implies \alpha\beta\gamma\leq\dfrac{32}{81}...(1)$ But in question it is given that $\alpha\beta\gamma=\dfrac{32}{81}$ and hence equality holds in $(1)$ and condition of equality gives $\alpha=2\beta=3\gamma$ . Using this and $\alpha\beta\gamma=\dfrac{32}{81}$ , we get $\alpha=4/3, \beta=2/3,\gamma=4/9$ , Now we want $b$ which is nothing but sum of products of roots $(\alpha,\beta,\gamma)$ taken two at a time i.e :- $\frac{4}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\dfrac 49+\dfrac 49\cdot\dfrac 43=\large\dfrac{16}{9}$

$\Large 5(9)-2(16)=\boxed{13}$