Problem 80 - Elementary and Intermediate Algebra

Question

Factor. $$ \frac{125}{8} s^{3}+\frac{1}{27} t^{3} $$

Step-by-Step Solution

Answer

\( \frac{125}{8} s^3 + \frac{1}{27} t^3 = \left(\frac{5}{2} s + \frac{1}{3} t\right) \left(\frac{25}{4} s^2 - \frac{5}{6} st + \frac{1}{9} t^2\right) \).

Step 1: Recognize the Structure

The given expression, \( \frac{125}{8} s^3 + \frac{1}{27} t^3 \), is a sum of two cubes. Recall that a sum of cubes can be factored using the formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).

Step 2: Identify Each Cube's Base

Notice that \( \left(\frac{5}{2} s\right)^3 \) is \( \frac{125}{8} s^3 \) and \( \left(\frac{1}{3} t\right)^3 \) is \( \frac{1}{27} t^3 \). Thus, we identify \( a = \frac{5}{2} s \) and \( b = \frac{1}{3} t \).

Step 3: Substitute into the Formula

Substitute \( a = \frac{5}{2} s \) and \( b = \frac{1}{3} t \) into the sum of cubes formula: \[ (a + b)(a^2 - ab + b^2) = \left(\frac{5}{2} s + \frac{1}{3} t\right) \left(\left(\frac{5}{2} s\right)^2 - \left(\frac{5}{2} s\right)\left(\frac{1}{3} t\right) + \left(\frac{1}{3} t\right)^2\right) \].

Step 4: Simplify Each Part

Simplify \( \left(\frac{5}{2} s\right)^2 = \frac{25}{4} s^2 \), \( \left(\frac{5}{2} s\right)\left(\frac{1}{3} t\right) = \frac{5}{6} st \), and \( \left(\frac{1}{3} t\right)^2 = \frac{1}{9} t^2 \).

Step 5: Write the Final Factored Form

Substitute back the simplified terms into the factored expression: \( \left(\frac{5}{2} s + \frac{1}{3} t\right)\left(\frac{25}{4} s^2 - \frac{5}{6} st + \frac{1}{9} t^2\right) \).

Key Concepts

Understanding the Sum of Cubes

When you encounter expressions like \( \frac{125}{8} s^3 + \frac{1}{27} t^3 \), you are dealing with a classic example of the **sum of cubes**. The sum of cubes is a special algebraic pattern that can be factored using a standard formula: \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\] Here’s a quick breakdown of how it works:

• Identify the cubes: Determine the two terms that are cubed in the expression. In our exercise, \( \left(\frac{5}{2} s\right)^3 \) and \( \left(\frac{1}{3} t\right)^3 \) fit this structure.
• Apply the formula: Once the terms \( a \) and \( b \) are recognized, substitute them into the formula above.
• Return the factored form: This results in a product of a binomial and a trinomial, giving the expression a simpler look.

Understanding this formula is crucial, as it turns a complicated expression into a more manageable form through theoretical structure.

Diving into Factoring Techniques

**Factoring techniques** are critical tools in algebra that simplify complex expressions, like transforming \( \frac{125}{8} s^3 + \frac{1}{27} t^3 \) into a more workable form. Here’s the approach applied to our problem: Identifying and applying the **sum of cubes formula** is one such technique. Here’s how we apply it:

• Express as cubes: Rewrite each term in the expression as a cube, i.e., \( \left(\frac{5}{2} s\right)^3 \) and \( \left(\frac{1}{3} t\right)^3 \).
• Substitute into the formula: Plug these base values into \((a + b)(a^2 - ab + b^2)\) to begin the factoring process.

This turns original polynomials into products of simpler expressions. Factoring is akin to peeling back layers of complexity to reveal components that are easier to manipulate. Factoring techniques, such as the sum of cubes, are not just mechanical tricks—they reveal the underlying structure of algebraic expressions.

Algebraic Expressions Demystified

**Algebraic expressions** like \( \frac{125}{8} s^3 + \frac{1}{27} t^3 \) represent numbers and operations in a symbolic form and are fundamental in mathematics. When manipulating these expressions: One vital aspect is the ability to **recognize patterns** and **apply formulas** effectively. Take our expression's handling as an example:

• Identify underlying structures: Such as recognizing the form of cubes which indicate that the sum of cubes might be a suitable technique.
• Manipulate symbols: Rewriting these as \( (a + b)(a^2 - ab + b^2) \) demonstrates manipulation skills, turning complex algebra into more approachable terms.

Breaking down algebraic expressions with pattern recognition not only aids in solving specific problems but also develops a more intuitive grasp of mathematics as a whole. Each expression, even as simple as the ones in this exercise, illustrates the beauty and precision of mathematical language and opens the door to broader problem-solving applications.