Neither of these worked fully, but they did offer some good things: the first eliminated the \(\sqrt[3]{2}\) term, while the second had \(\sqrt[3]{2}\times \sqrt[3]{4}=2\).

One thing we could do is to combine them and use both \(\sqrt[3]{2}\)and\(\sqrt[3]{4}\). As we don’t know how many of each to use, we will try multiplying by \(1+a\sqrt[3]{2}+b\sqrt[3]{4}\) and see where the algebra takes us. (If it doesn’t work, we could always then try replacing \(1\) by a variable to see whether that helps.)

We have
\[\begin{align*}
(1+\sqrt[3]{2})(1+a\sqrt[3]{2}+b\sqrt[3]{4}) &= 1+a\sqrt[3]{2}+b\sqrt[3]{4} + \sqrt[3]{2}+a\sqrt[3]{4}+2b\\
&= (1+2b)+(1+a)\sqrt[3]{2}+(a+b)\sqrt[3]{4}.
\end{align*}\]

So if we take \(a=-1\) and \(b=1\), both of the cube root terms will disappear, and we will be left with \(1+2b=3\).

So to rationalise our fraction, we multiply both the numerator and denominator by \(1-\sqrt[3]{2}+\sqrt[3]{4}\) to find that \[\dfrac{1}{1+\sqrt[3]{2}}=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 3}.\]

When we rationalise a denominator with a square root such as \(1+\sqrt{2}\), we multiply it by \(1-\sqrt{2}\), since we have the algebraic identity \[a^2-b^2=(a+b)(a-b),\] the well-known difference of two squares identity.

There are similar identities for the sum and difference of two cubes:
\[\begin{align*}
a^3+b^3&=(a+b)(a^2-ab+b^2)\\
a^3-b^3&=(a-b)(a^2+ab+b^2).
\end{align*}\]

So in our case, we are starting with \(1+\sqrt[3]{2}\), so if we use the first identity, we find that \[(1+\sqrt[3]{2})(1-\sqrt[3]{2}+\sqrt[3]{4}) = 1^3+(\sqrt[3]{2})^3=3,\] giving \[\dfrac{1}{1+\sqrt[3]{2}}=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 3}\] as before.

You might wonder why there isn’t a similar identity for \(a^2+b^2\) to parallel that for \(a^3+b^3\).

If \(a^3+b^3=0\), then \(a^3=-b^3\), so \(a=-b\) is a possibility. This means that \(a+b\) is a factor of \(a^3+b^3\) (using the factor theorem).

However, if \(a^2+b^2=0\), then \(a^2=-b^2\), which we cannot square root in the reals (because \(-1\) does not have a real square root). If, however, we allow ourselves to use complex numbers, then we get \(a=\pm ib\) (where \(i^2=-1\)), so \(a\pm ib\) are factors of \(a^2+b^2\). Thus we can write \[a^2+b^2=(a+ib)(a-ib),\] which turns out to be a very useful identity when working with complex numbers.

As discussed in the Suggestion section, we could find a polynomial satisfied by \(x=1+\sqrt[3]{2}\), and then rearrange it in order to find a number we could multiply \(x\) by to obtain an integer.

We’ll look at the methods from the Irrational roots solution for finding a polynomial satisfied by \(1+\sqrt{2}\), and see whether we can adapt them to our case.

Irrational roots approach 1

This looks like it will be pretty horrendous in our context, so we won’t try using it here!

Irrational roots approach 2

This involved rearranging \(x=1+\sqrt{2}\) to isolate the square root. We could do the same here.

If \(x=1+\sqrt[3]{2}\), then \(x-1=\sqrt[3]{2}\).

We can now cube this to get \((x-1)^3=2\).

Expanding gives \[x^3-3x^2+3x-1=2,\] so \[x^3-3x^2+3x=3.\] So \(x(x^2-3x+3)=3\), which is what we needed.

Thus if we multiply the denominator of our fraction, \(x=1+\sqrt[3]{2}\), by
\[\begin{align*}
x^2-3x+3&=(1+\sqrt[3]{2})^2-3(1+\sqrt[3]{2})+3\\
&=1+2\sqrt[3]{2}+\sqrt[3]{4}-3-3\sqrt[3]{2}+3\\
&=1-\sqrt[3]{2}+\sqrt[3]{4}
\end{align*}\]

we will get \(3\), hence \[\frac{1}{1+\sqrt[3]{2}}=\frac{1-\sqrt[3]{2}+\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 3}.\]

Irrational roots approach 3

This involved translating a graph. We want a graph which has a root at \(x=1+\sqrt[3]{2}\). If we begin with the graph of \(y=x^3-2\), which has a root at \(\sqrt[3]{2}\), and then translate it \(1\) unit to the right, we end up with a graph which has a root at \(x=1+\sqrt[3]{2}\).

So the graph \(y=(x-1)^3-2\) has a root at \(1+\sqrt[3]{2}\), and our required polynomial is \((x-1)^3-2=x^3-3x^2+3x-3\).

We can then continue as above to find that \(x(x^2-3x+3)=3\), and hence to rationalise the denominator of our fraction.

Irrational roots approach 4

This involved squaring \(x=1+\sqrt{2}\) and then eliminating \(\sqrt{2}\). We could try the same here.

We can combine the \(x^3\) and \(x^2\) equations to eliminate the \(\sqrt[3]{4}\) terms: \[x^3-3x^2=-3\sqrt[3]{2}.\] Then we can use the \(x\) equation to eliminate the \(\sqrt[3]{2}\) term from this: \[x^3-3x^2+3x=3,\] and we have our required polynomial equation.

The rest continues as in “Irrational roots approach 2”.

Which of these approaches do you prefer?

Which of these approaches can be most easily adapted for the other problems or other situations?

For the remaining questions, we will only offer one approach to solving them, but as we have seen, there are several other possibilities.

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

This looks very similar to the first problem. We’ll use the algebraic identity for \(a^3+b^3\) to answer this one.

We have \[2^3+(\sqrt[3]{2})^3=(2+\sqrt[3]{2})(4-2\sqrt[3]{2}+\sqrt[3]{4})\] and the left hand side equals \(10\), so \[\frac{1}{2+\sqrt[3]{2}} = \frac{4-2\sqrt[3]{2}+\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 10}.\]

\(\dfrac{1}{\sqrt[3]{2} + \sqrt[3]{4}}\)

In this case, we notice that the denominator has a factor of \(\sqrt[3]{2}\), so we can write \[\dfrac{1}{\sqrt[3]{2} + \sqrt[3]{4}}=\dfrac{1}{\sqrt[3]{2}(1 + \sqrt[3]{2})}.\] We can rationalise the second factor using our answer to question 1, and we can multiply the numerator and denominator by \(\sqrt[3]{4}\) to rationalise the \(\sqrt[3]{2}\) term as in the warm-up problem, giving: \[\dfrac{1}{\sqrt[3]{2}(1 + \sqrt[3]{2})}=\frac{\sqrt[3]{4}}{2}\times \frac{1-\sqrt[3]{2}+\sqrt[3]{4}}{3}=
\frac{\sqrt[3]{4}-2+2\sqrt[3]{2}}{\vphantom{\sqrt[3]{2}} 6}.\]

\(\dfrac{1}{1 - \sqrt[3]{2} + \sqrt[3]{4}}\)

We saw in our answer to question 1 above that \[(1+\sqrt[3]{2})(1-\sqrt[3]{2}+\sqrt[3]{4})=3.\] So just as \(\dfrac{1}{1+\sqrt[3]{2}}=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 3}\), we have \[\dfrac{1}{1-\sqrt[3]{2}+\sqrt[3]{4}}=\dfrac{1+\sqrt[3]{2}}{\vphantom{\sqrt[3]{2}} 3}.\]

Taking it further: What difference does a sign make?

\(\dfrac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}\)

In approach 2 (algebraic identities) to the first problem above, we saw two useful identities. One of them was \[a^3-b^3=(a-b)(a^2+ab+b^2).\] If we take \(a=\sqrt[3]{2}\) and \(b=1\), we obtain \[2-1=(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1),\] so that \[\dfrac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\sqrt[3]{2}-1,\] and we don’t even have a fraction left!

\(\dfrac{1}{1 + \sqrt[3]{2} - \sqrt[3]{4}}\)

In this case, there are no algebraic identities which are obviously useful, because the signs do not match the ones we know. So we’ll use one of the polynomial equation approaches for this instead. It is also not obviously a translation of a simple graph, so we will make use of the fourth polynomial equation approach.

We have (with a bit of algebra, and using the facts that \((\sqrt[3]{2})^2=\sqrt[3]{4}\), \((\sqrt[3]{4})^2=2\sqrt[3]{2}\) and \(\sqrt[3]{2}\times \sqrt[3]{4}=2\)):
\[\begin{align*}
x &= 1 + \sqrt[3]{2} - \sqrt[3]{4}\\
x^2 &= -3 + 4\sqrt[3]{2} - \sqrt[3]{4}\\
x^3 &= -13 + 3\sqrt[3]{2} + 6\sqrt[3]{4}.
\end{align*}\]
We can eliminate the \(\sqrt[3]{4}\) term from these equations to get expressions only involving \(\sqrt[3]{2}\), for example:
\[\begin{align*}
x^3 + 6x &= -7 + 9\sqrt[3]{2}\\
x^2 - x &= -4 + 3\sqrt[3]{2}.
\end{align*}\]

Then we can eliminate \(\sqrt[3]{2}\) by subtracting \(3\) times the second of these two equations from the first to give: \[(x^3+6x)-3(x^2-x) = 5,\] so that \(x^3-3x^2+9x=5\), giving \(x(x^2-3x+9)=5\).

Thus we wish to multiply the numerator and denominator by \[x^2-3x+9=(-3+4\sqrt[3]{2}-\sqrt[3]{4})-3(1+\sqrt[3]{2}-\sqrt[3]{4})+9=3+\sqrt[3]{2}+2\sqrt[3]{4}\] to get \[\frac{1}{1+\sqrt[3]{2}-\sqrt[3]{4}} = \frac{3+\sqrt[3]{2}+2\sqrt[3]{4}}{\vphantom{\sqrt[3]{2}} 5}.\]

The point of these problems is not so much to work out the answers to these particular questions, but rather to show that we can, in principle, work out the answers. Computer algebra systems which do these calculations use techniques such as these behind the scenes to do their work.

Can you find any fractions for which you cannot rationalise the denominator?

The approach we have taken has involved finding a polynomial with the denominator as a root. There are numbers which are not the root of any polynomial with integer coefficients, such as \(1+\pi\) (see Squaring the circle), so a fraction such as \(\dfrac{1}{1+\pi}\) cannot have its denominator rationalised.