pls helpppp

That simplifies to 2*3^(1/3).  Hope this helps!

Whenever there are multiple radical terms in the denominator, the process of simplification becomes quite a challenge, but I realized after some time that the denominator $1 - \sqrt[3]{3} + \sqrt[3]{9} = \left(\sqrt[3]{3}\right)^2 - \sqrt[3]{3} + 1$ looks to take the form of the beginnings of the factorization of the sum of cubes.

In the general factorization, $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. If we consider the case where $a = \sqrt[3]{3} \text{ and } b = 1$, then we can multiply the denominator by $\sqrt[3]{3} + 1$ to transform the denominator into a sum of cubes and then simplify. 

$\frac{4}{1 - \sqrt[3]3 + \sqrt[3]9} * \frac{\sqrt[3]{3} + 1}{\sqrt[3]{3} + 1} = \frac{4\left(\sqrt[3]{3} + 1\right)}{\left(\sqrt[3]{3}\right)^3 + 1^3} = \frac{4\left(\sqrt[3]{3} + 1\right)}{3 + 1} = \sqrt[3]{3} + 1$