Rationalizing a quadratic surd $a+b\sqrt{d}$ entails multiplying it by its conjugate. In this case, the word "conjugate" is as simple as replacing a binomial sum $A+B$ with a difference $A-B$ and vice versa. However, the word "conjugate" has a much more advanced generalization in the contexts of field theory and Galois theory.
Suppose that $K$ is a field (like $\mathbb{Q},\mathbb{R},\mathbb{C}$), and $f(x)$ is a polynomial in the variable $x$ with coefficients from $K$. If $f(x)$ is irreducible (cannot be factored using other polynomial also with coefficients from $K$) and $\alpha$ is a root of $f(x)$ (so $\alpha$ may not be in $K$, it may be in a bigger field $L$ containing $K$), then the nontrivial conjugates of $\alpha$ are all the other roots of $f(x)$.
If $L$ is a sufficiently nice "extension" of $K$, there are a set of "symmetries" of $L$ (functions $L\to L$) which do nothing to the elements of $K$, preserve the arithmetic operations of addition and multiplication on $L$, and in particular it follows that they permute the roots of $f(x)$. Indeed, they will permute them "transitively," meaning given any two roots $\alpha,\alpha'$ there is a "symmetry" $\sigma:L\to L$ for which $\sigma(\alpha)=\alpha'$.
Let's see this concretely. Say $K=\mathbb{Q}$ and $f(x)=x^2+1$. One root is $\alpha=i$, the other is $\alpha'=-i$, and there is a "symmetry" $\sigma:\mathbb{C}\to\mathbb{C}$ which does nothing to real numbers and $\sigma(i)=-i$. In fact, this is just complex conjugation $\sigma(a+bi)=a-bi$. It "preserves the operations" of addition and subtraction in the sense that $\sigma(z+w)=\sigma(z)+\sigma(w)$ and $\sigma(zw)=\sigma(z)\sigma(w)$ for all $z,w\in\mathbb{C}$.
One thing to notice is that $\sigma(r)=r$ is not only true for $r\in\mathbb{R}$, it is only true for real numbers $r$.
Another example. Let $K=\mathbb{Q}$ and $f(x)=x^2-2$. The roots are $\alpha=\sqrt{2}$ and $\alpha'=-\sqrt{2}$. These exist in the bigger field $L=\mathbb{Q}(\sqrt{2})$. (By definition this means "the smallest field containing $\sqrt{2}$ and $\mathbb{Q}$ within $\mathbb{C}$," but this happens to be the same as $\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$, which you can prove as an exercise). Once again there is a nontrivial symmetry $\sigma:\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{2})$ which does nothing to rationals, preserves operations, and $\sigma(\sqrt{2})=-\sqrt{2}$. The function is given by the "conjugate," as in $\sigma(a+b\sqrt{2})=a-b\sqrt{2}$; you can check by hand this satisfies the properties $\sigma(x+y)=\sigma(x)+\sigma(y)$ and $\sigma(xy)=\sigma(x)\sigma(y)$ for all $x=a+b\sqrt{d}$, $y=p+q\sqrt{d}$.
Now let $K=\mathbb{Q}$ and $f(x)$ be any irreducible rational-coefficient polynomial, $\alpha$ a root, and $L$ the smallest subfield of $\mathbb{C}$ containing all of the roots of $\mathbb{Q}$. (Note this may be bigger than $\mathbb{Q}(\alpha)$; there may be some other roots $\alpha'$ of $f(x)$ that cannot be expressed in terms of $\alpha$ using rational functions as every element in $\mathbb{Q}(\alpha)$ is.) There may be more than one symmetry $\sigma$, there could be multiple. In any case, if we apply all of the symmetries $\sigma_1,\sigma_2,\cdots$ to $\alpha$ we get all of the roots $\alpha_1,\alpha_2,\cdots$ of the polynomial $f(x)$. Then if we multiply them all together, by Vieta's formulas we find the product of all the zeros $\alpha_1\cdots\alpha_n$ is $\pm$ the constant coefficient of $f(x)$, in particular the product is rational.
That is: the product of an algebraic number and all of its conjugates will be rational.
Let's apply this with $f(x)=x^3-d$, where $d$ is not a perfect cube. Without loss of generality, $d$ is just an integer. The roots are $\sqrt[3]{d},~\omega\sqrt[3]{d}$, and $\omega^2\sqrt[3]{d}$, where $\sqrt[3]{d}$ is the real cube root of $d$ and $1,\omega,\omega^2$ are the three cube roots of unity. In particular, $\omega=\exp(2\pi i/3)$.
The two nontrivial symmetries $\sigma_1$ and $\sigma_2$ satisfy $\sigma_1(\sqrt[3]{d})=\omega\sqrt[3]{d}$ and $\sigma_2(\sqrt[3]{d})=\omega^2\sqrt[3]{d}$, so
$$\begin{array}{ll} \sigma_1(a+b\sqrt[3]{d}+c\sqrt[3]{d^2}) & =a+b\sigma_1(\sqrt[3]{d})+\sigma_1(\sqrt[3]{d})^2 \\ & = a+b\omega\sqrt[3]{d}+b\omega^2\sqrt[3]{d^2} \end{array}$$
$$\begin{array}{ll} \sigma_2(a+b\sqrt[3]{d}+c\sqrt[3]{d^2}) & =a+b\sigma_2(\sqrt[3]{d})+c\sigma_2(\sqrt[3]{d})^2 \\ & = a+b\omega^2\sqrt[3]{d}+\omega\sqrt[3]{d^2} \end{array} $$
because $\sigma_{1,2}$ preserve addition and multiplication and $(\omega^2)^2=\omega$ at the end. Of course the trivial symmetry $\sigma_0$ doesn't do anything, $\sigma_0(a+b\sqrt[3]{d}+c\sqrt[3]{d^2})=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}$.
Thus to rationalize $x=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}$, one multiplies it by its conjugates $\sigma_1(x)$ and $\sigma_2(x)$.
While $\sigma_1(x)$ and $\sigma_2(x)$ may not be real numbers, the product $\sigma_0(x)\sigma_1(x)\sigma_2(x)$ will be rational, and further since $\sigma_0(x)$ is rational, the product $\sigma_1(x)\sigma_2(x)$ will be rational as well, which is what we would multiply the numerator by if $x$ was in the denominator.
(I make no claim this is the easiest or most efficient way of rationalizing the denominator - I'm sure it isn't. I just wanted to showcase some Galois theory.)