The given mathematical expression is:

$\frac{9}{1 + \sqrt[3]{2}}$

To simplify this, we rationalize the denominator by multiplying both numerator and denominator by the conjugate of $1 + \sqrt[3]{2}$, which is $1 - \sqrt[3]{2} + (\sqrt[3]{2})^2$. This approach leverages the fact that the cube of a binomial will simplify due to the formula:

$(a + b)(a^2 - ab + b^2) = a^3 + b^3$

Here are the steps:

1. Multiply numerator and denominator by $1 - \sqrt[3]{2} + (\sqrt[3]{2})^2$: [ \frac{9}{1 + \sqrt[3]{2}} \cdot \frac{1 - \sqrt[3]{2} + (\sqrt[3]{2})^2}{1 - \sqrt[3]{2} + (\sqrt[3]{2})^2} ]

2. The denominator becomes: [ (1 + \sqrt[3]{2})(1 - \sqrt[3]{2} + (\sqrt[3]{2})^2) = 1^3 - (\sqrt[3]{2})^3 = 1 - 2 = -1 ]

3. The numerator becomes: [ 9 \cdot (1 - \sqrt[3]{2} + (\sqrt[3]{2})^2) = 9 - 9\sqrt[3]{2} + 9(\sqrt[3]{2})^2 ]

Thus, the expression simplifies to: [

• (9 - 9\sqrt[3]{2} + 9(\sqrt[3]{2})^2) ]

Or equivalently: [ -9 + 9\sqrt[3]{2} - 9(\sqrt[3]{2})^2 ]

Let me know if you'd like further breakdowns or explanations!

Related Questions

1. What are the advantages of rationalizing cube roots?
2. How does the cube expansion formula work in other scenarios?
3. Can you provide examples with higher-order roots?
4. How is this approach similar or different from rationalizing square roots?
5. How does the expression change if the denominator was $1 - \sqrt[3]{2}$?

Tip

Rationalizing denominators simplifies expressions and helps avoid approximations, especially when solving equations or working symbolically!