Let $a$ and $b$ be real numbers such that $a^2b^3=\frac{32}{27}$ and $\frac{a}{b^3}=\frac{27}{4}$. What is $a+b$?

Question

Let $a$ and $b$ be real numbers such that $a^2b^3=\frac{32}{27}$ and $\frac{a}{b^3}=\frac{27}{4}$.  What is $a+b$?

GPT 3.5 · Answer

Multiplying the second given equation by $b^3$ gives that $a=b^3\cdot\frac{27}{4}$. Substituting this into the first given equation gives $(b^3)^2b^3\cdot\frac{27}{4}=\frac{32}{27}$, which simplifies to $b^9=\frac{32}{27}\cdot\frac{4}{27}=\frac{128}{243}$. We have $b^3=\sqrt[9]{b^9}=\sqrt[9]{\frac{128}{243}}=\frac{2}{3}$. This gives $a=b^3\cdot\frac{27}{4}=\frac{2}{3}\cdot\frac{27}{4}=\frac{27}{6}=\frac{9}{2}$. Therefore $a+b=\frac{9}{2}+\frac{2}{3}=\frac{27+4}{6}=\boxed{\frac{31}{6}}$.