Can you rationalise the denominator of the following fractions?

\(\dfrac{1}{1+\sqrt[3]{2}}\)

There are several ways one could approach this problem.

We know that when rationalising a denominator such as \(1+\sqrt{2}\), we multiply by \(1-\sqrt{2}\). So what would happen if we multiplied our denominator by \(1-\sqrt[3]{2}\)? What about \(1-\sqrt[3]{4}\)? Perhaps we can tweak these somehow to find something we can multiply by to get rid of the roots?

When we rationalise a denominator such as \(1+\sqrt{2}\), we make use of the difference of two squares. Is there some similar identity we could use in our case, perhaps one involving two cubes?

The problem Irrational roots asked us to find a quadratic polynomial with integer coefficients which has \(x = 1+\sqrt{2}\) as a root. We found that \(x^2-2x-1=0\) does the job, and we can rearrange this answer to give \(x(x-2)=1\). There are various approaches to solving that problem offered in the solution. Can we apply these methods to find a polynomial with integer coefficients which has our denominator, \(x=1+\sqrt[3]{2}\), as a root? How will this help us?