Chapter 6: Problem 80
Factor. $$ \frac{125}{8} s^{3}+\frac{1}{27} t^{3} $$
Short Answer
Expert verified
\( \frac{125}{8} s^3 + \frac{1}{27} t^3 = \left(\frac{5}{2} s + \frac{1}{3} t\right) \left(\frac{25}{4} s^2 - \frac{5}{6} st + \frac{1}{9} t^2\right) \).
Step by step solution
01
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: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).
02
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 \( b = \frac{1}{3} t \).
03
Substitute into the Formula
Substitute \( a = \frac{5}{2} s \) and \( b = \frac{1}{3} t \) into the sum of cubes formula: \[ (a + b)(a^2 - ab + b^2) = \left(\frac{5}{2} s + \frac{1}{3} t\right) \left(\left(\frac{5}{2} s\right)^2 - \left(\frac{5}{2} s\right)\left(\frac{1}{3} t\right) + \left(\frac{1}{3} t\right)^2\right) \].
04
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 \( \left(\frac{1}{3} t\right)^2 = \frac{1}{9} t^2 \).
05
Write the Final Factored Form
Substitute back the simplified terms into the factored expression: \( \left(\frac{5}{2} s + \frac{1}{3} t\right)\left(\frac{25}{4} s^2 - \frac{5}{6} st + \frac{1}{9} t^2\right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.