Algebra

\( \frac{4}{1 - \sqrt[3]{3}} + \sqrt[3]{9} \)

Note that   1 - 3   can  be written as  ( 1 - 3^(1/3)) ( 1 + 3^(1/3) + 3^(2/3))

Multiply num/den  of the first fraction by the expression in re and  we  get

4 (1 + 3^(1/3)  + 3^(2/3) )

_____________________ =   -2 ( 1 + 3^(1/3) + 3^(2/3) )  =   -2 - 2*3^(1/3) -2*3^(2/3)

          1 - 3

And   9^(1/3)  =   3^(2/3)

So we have

-2 -2*3^(1/3) - 2 *3^(2/3) + 3^(2/3)  =

-2 - 2*3^(1/3) - 3^(2/3)