How to Rationalize a Denominator with Four Terms

To rationalize

$$\frac{1}{1+\sqrt{12}-\sqrt[3]{3}-\sqrt[3]{9}},$$

remove radicals from the denominator by using two identities in sequence. The square root is handled with a conjugate; the cube roots are handled using $(\sqrt[3]{3})^3=3$.

The result is

$$\frac{-2+\sqrt[3]{3}+5\sqrt[3]{9}+7\sqrt{3}+2\sqrt{3}\sqrt[3]{3}-\sqrt{3}\sqrt[3]{9}}{22}.$$

The denominator is now the rational number $22$. The numerator still contains radicals, which is normal: rationalizing a denominator does not mean eliminating radicals from the entire expression.

Step 1: Rewrite the radicals

Let

$$a=\sqrt{3}, \qquad u=\sqrt[3]{3}.$$

Then $\sqrt{12}=2\sqrt{3}=2a$, and $\sqrt[3]{9}=u^2$, because $u^2=\sqrt[3]{3^2}$. The denominator becomes

$$1+2a-u-u^2.$$

Group the terms without $a$ by setting $x=1-u-u^2$. The denominator is $x+2a$, whose conjugate with respect to the square root is $x-2a$.

Step 2: Clear the square root

Multiply the fraction by $(x-2a)/(x-2a)$. Since $a^2=3$, the new denominator is

$$(x+2a)(x-2a)=x^2-4a^2=x^2-12.$$

Expand $x^2$ and use $u^3=3$ and $u^4=3u$:

$$\begin{aligned} x^2&=(1-u-u^2)^2\\ &=1-2u-2u^2+u^2+2u^3+u^4\\ &=1-2u-u^2+2(3)+3u\\ &=7+u-u^2. \end{aligned}$$

Therefore $x^2-12=u-u^2-5$, and the fraction is now

$$\frac{1-u-u^2-2a}{u-u^2-5}.$$

Step 3: Clear the cube roots

The denominator still contains $u$ and $u^2$. Use $u^3=3$ to check the identity

$$(u-u^2-5)(u^2-2u-7)=44.$$

For example, expand the product and replace $u^3$ with $3$ and $u^4$ with $3u$; all terms involving $u$ cancel. Multiply the numerator and denominator by $u^2-2u-7$. The denominator is now $44$.

For the numerator, first multiply the parts that do not contain $a$:

$$(u^2-2u-7)(1-u-u^2)=-4+2u+10u^2.$$

The part containing $a$ is

$$-2a(u^2-2u-7)=14a+4au-2au^2.$$

So the fraction becomes

$$\frac{-4+2u+10u^2+14a+4au-2au^2}{44}.$$

Divide the numerator and denominator by $2$, then substitute $a=\sqrt{3}$ and $u=\sqrt[3]{3}$, with $u^2=\sqrt[3]{9}$. This gives the stated result.

Why the two multipliers work

Each multiplier matches a different kind of radical. The first is the ordinary conjugate $x-2a$, which uses the difference-of-squares identity and replaces $(x+2a)(x-2a)$ with $x^2-12$. The second is chosen for the remaining expression $u-u^2-5$; its product with $u^2-2u-7$ is rational because powers of $u$ reduce using $u^3=3$.

This method is more reliable than trying to guess one large multiplier for the original four-term denominator. It also keeps the arithmetic manageable: simplify the square-root part first, then handle the cube-root terms.

Common mistakes

• Treating $\sqrt[3]{9}$ as $3\sqrt[3]{3}$. In fact, $\sqrt[3]{9}=u^2$, not $3u$.
• Using the conjugate of only part of the denominator and stopping. The first conjugate removes the square root from that product, but leaves cube roots in the denominator; a second step is needed.
• Dropping terms while expanding $(1-u-u^2)^2$. The cross term $2u^3$ and the square term $u^4$ both matter. Reduce powers only after writing the full expansion.
• Expecting a rational final value. The goal is a rational denominator, not a rational whole expression. The original number is irrational, so its numerator generally retains radicals.

As a quick numerical check, the original denominator is about $0.942$, so its reciprocal is about $1.062$. The rationalized expression has the same approximate value.