You need to remove the radicals from denominator, hence you need to remember the following formula such that:

`x^3 + y^3 = (x + y)(x^2 - xy + y^2)`

Notice that denominator only contains the `(x+y)` part and you need to multiply by `(x^2 - xy + y^2)` part to remove...

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You need to remove the radicals from denominator, hence you need to remember the following formula such that:

`x^3 + y^3 = (x + y)(x^2 - xy + y^2)`

Notice that denominator only contains the `(x+y)` part and you need to multiply by `(x^2 - xy + y^2)` part to remove the cube root.

You need to substitute `2root(3)a` for x and `3root(3)b` for y and you need to multiply both numerator and denominator by `(4root(3)(a^2) - 6root(3)(ab) + 9root(3)(b^2))` such that:

`(4root(3)(a^2) - 6root(3)(ab) + 9root(3)(b^2))/((2root(3)a + 3root(3)b)(4root(3)(a^2)- 6root(3)(ab) + 9root(3)(b^2))) = (4root(3)(a^2) - 6root(3)(ab) + 9root(3)(b^2))/(8a+ 27b)`

**Hence, rationalizing the fraction yields `1/(2root(3)a + 3root(3)b) = (4root(3)(a^2) - 6root(3)(ab) + 9root(3)(b^2))/(8a + 27b).` **