rationalise\:\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}

To rationalize the expression \(\frac{1}{\sqrt[3]{a} - \sqrt[3]{b}}\), we need to eliminate the cube root in the denominator. The rationalization involves multiplying both the numerator and the denominator by the conjugate of the denominator. 

For cube roots, the conjugate is obtained by using the identity:

\[
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
\]

So, the conjugate of \(\sqrt[3]{a} - \sqrt[3]{b}\) is \(\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}\).

### Step-by-step Solution:

1. **Multiply by the Conjugate:**

\[
\frac{1}{\sqrt[3]{a} - \sqrt[3]{b}} \times \frac{\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}}{\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}}
\]

2. **Simplify the Denominator:**

\[
(\sqrt[3]{a} - \sqrt[3]{b})(\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}) = (\sqrt[3]{a})^3 - (\sqrt[3]{b})^3 = a - b
\]

Thus, the denominator simplifies to \(a - b\).

3. **Final Expression:**

\[
\frac{\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}}{a - b}
\]

### Rationalized Form:

\[
\frac{\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}}{a - b}
\]

This is the rationalized form of the expression \(\frac{1}{\sqrt[3]{a} - \sqrt[3]{b}}\).

Would you like a more detailed explanation or have any questions?

### Related Questions:

1. How do you rationalize expressions with square roots in the denominator?
2. Can you explain the difference between rationalizing a cube root vs. a square root?
3. What is the conjugate of a binomial involving cube roots?
4. How do you simplify complex fractions involving roots?
5. What are other methods for simplifying expressions with roots?

**Tip:** Always remember that rationalizing the denominator involves removing any roots or irrational numbers from the denominator by multiplying by a suitable conjugate.

rationalise:\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}

Learn how to rationalize the expression 1/(cube root of a - cube root of b) with a detailed step-by-step solution. This problem involves algebraic concepts such as rationalization, cube roots, and the difference of cubes theorem. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation