Problem 42 - Intermediate Algebra

Question

Add or subtract. $$ \frac{\sqrt[3]{3}}{10}+\sqrt[3]{\frac{24}{125}} $$

Step-by-Step Solution

Answer

\( \frac{\sqrt[3]{3}}{2} \)

Step 1: Simplify the second term

First consider the expression \( \sqrt[3]{\frac{24}{125}} \). Recognize this as a cube root of a fraction. The term can be split into \( \frac{\sqrt[3]{24}}{\sqrt[3]{125}} \). \( \sqrt[3]{125} \) simplifies to 5 since \( 5^3 = 125 \).

Step 2: Simplify \( \sqrt[3]{24} \)

To express \( \sqrt[3]{24} \) in a simpler form, note that 24 can be factored into \( 8 \times 3 \). Thus, we have \( \sqrt[3]{24} = \sqrt[3]{8 \times 3} = \sqrt[3]{8} \times \sqrt[3]{3} \). Since \( \sqrt[3]{8} = 2 \) because \( 2^3 = 8 \), this gives us \( 2 \sqrt[3]{3} \).

Step 3: Substitute the simplified form back in

Replace \( \frac{\sqrt[3]{24}}{\sqrt[3]{125}} \) with \( \frac{2 \sqrt[3]{3}}{5} \). Now, the expression is \( \frac{\sqrt[3]{3}}{10} + \frac{2 \sqrt[3]{3}}{5} \).

Step 4: Find common denominator and add fractions

To add the fractions, find a common denominator. The least common multiple of 10 and 5 is 10. The first term already has a denominator of 10. Multiply the numerator and denominator of the second term by 2 to achieve the common denominator: \( \frac{2 \sqrt[3]{3}}{5} = \frac{4 \sqrt[3]{3}}{10} \).

Step 5: Add the fractions

Now add the two fractions: \( \frac{\sqrt[3]{3}}{10} + \frac{4 \sqrt[3]{3}}{10} = \frac{(1 + 4) \sqrt[3]{3}}{10} = \frac{5 \sqrt[3]{3}}{10} \).

Step 6: Simplify the result

Simplify \( \frac{5 \sqrt[3]{3}}{10} \) by dividing the numerator and the denominator by 5 to get \( \frac{\sqrt[3]{3}}{2} \).

Key Concepts

Simplifying Radicals

Radicals can look intimidating, but simplifying them helps us make expressions more manageable. For cube roots, like \( \sqrt[3]{8} \), we aim to express the number under the root in the simplest form. Breaking it into factors that are perfect cubes is key. For instance, we know \( 8 = 2^3 \) so \( \sqrt[3]{8} = 2 \), simplifying the expression immediately.

This process also applies when handling cube roots of a multiplication, such as \( \sqrt[3]{24} \). Split 24 into its factors and look for any cubed numbers. Since \( 8 \times 3 = 24 \) and \( \sqrt[3]{8} = 2 \), it simplifies to \( 2 \times \sqrt[3]{3} \). This process reduces complexity and makes it easier to work with radical expressions.

Cube Roots

Cube roots are a special type of radical expression, represented as \( \sqrt[3]{x} \). They are the number that, when multiplied by itself three times, gives the original number \( x \). For example, \( \sqrt[3]{125} \) equals 5 because \( 5^3 = 125 \).

When working with cube roots of fractions, you can simplify them by taking the cube root of the numerator and the denominator separately. As in \( \sqrt[3]{\frac{24}{125}} \), it can be broken down to \( \frac{\sqrt[3]{24}}{\sqrt[3]{125}} \). This separation facilitates simplifying each part individually.

Adding Fractions

Adding fractions requires having a common denominator. This means both fractions must have the same bottom number. It allows us to add the numerators directly while maintaining the denominator intact, simplifying the addition process. For example, adding:

• \( \frac{3}{5} + \frac{2}{5} = \frac{3 + 2}{5} = \frac{5}{5} = 1 \)

If fractions have different denominators, finding the common denominator becomes necessary. This ensures that all parts of the fractions are on similar terms, allowing for straightforward addition of numerators.

Finding Common Denominator

To combine fractions like \( \frac{\sqrt[3]{3}}{10} + \frac{2\sqrt[3]{3}}{5} \), compute their least common denominator (LCD). This means identifying the smallest multiple common to both denominators. Here, the LCD for 10 and 5 is 10 since 10 is already a multiple of 5.

This solution requires adjusting only the second fraction to match the first. By multiplying both the numerator and denominator by 2, \( \frac{2\sqrt[3]{3}}{5} \) becomes \( \frac{4\sqrt[3]{3}}{10} \), making it compatible for addition without changing its value.

Once the fractions have the same denominator, adding them becomes a simple arithmetic task.