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 $a+b$?
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 $a+b=\frac{9}{2}+\frac{2}{3}=\frac{27+4}{6}=\boxed{\frac{31}{6}}$.
To solve this problem, we need to find the values of $a$ and $b$ that satisfy the given equations and then find the value of $a+b$.
Let's start by solving the first equation: $a^2b^3=\frac{32}{27}$.
We can rewrite this equation as $(ab)^3=\frac{32}{27}$. Taking the cube root on both sides, we get $ab=\left(\frac{32}{27}\right)^{\frac{1}{3}}$.
Now, let's move on to the second equation: $\frac{a}{b^3}=\frac{27}{4}$.
We can rearrange this equation to get $a=\frac{27}{4}b^3$.
Now, substitute the value of $a$ from the second equation into the first equation:
$ab=\left(\frac{32}{27}\right)^{\frac{1}{3}}$
$\frac{27}{4}b^3 \cdot b=\left(\frac{32}{27}\right)^{\frac{1}{3}}$
Simplifying, we have:
$\frac{27}{4}b^4=\left(\frac{32}{27}\right)^{\frac{1}{3}}$
Now, let's solve for $b$:
$b^4=\frac{\left(\frac{32}{27}\right)^{\frac{1}{3}}}{\frac{27}{4}}$
$b^4=\frac{4}{27}\left(\frac{32}{27}\right)^{\frac{1}{3}}$
Simplifying further, we get:
$b^4=\frac{4}{27}\cdot\frac{2}{3}$
$b^4=\frac{8}{81}$
Taking the fourth root of both sides, we have:
$b=\left(\frac{8}{81}\right)^{\frac{1}{4}}$
Simplifying, we get:
$b=\frac{2}{3}$
Now, substitute the value of $b$ back into the second equation to find the value of $a$:
$a=\frac{27}{4}\left(\frac{2}{3}\right)^3$
$a=\frac{27}{4}\cdot\frac{8}{27}$
Simplifying, we get:
$a=2$
Finally, we can find $a+b$:
$a+b=2+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}$.
Therefore, $a+b$ is equal to $\frac{8}{3}$.