/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Factor the polynomial. $$a^{6}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 27

Factor the polynomial. $$a^{6}-b^{6}$$

Short Answer

Expert verified
The polynomial factors to \((a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2)\).

Step by step solution

01

Recognize the Difference of Squares

The given polynomial is \( a^6 - b^6 \), which is in the form \( x^2 - y^2 \). This can be factored using the difference of squares formula: \( x^2 - y^2 = (x - y)(x + y) \).
02

Apply the Difference of Squares

Apply the difference of squares: \( a^6 - b^6 = (a^3)^2 - (b^3)^2 = (a^3 - b^3)(a^3 + b^3) \).
03

Recognize the Sum and Difference of Cubes

The expression consists of two terms: \( a^3 - b^3 \) and \( a^3 + b^3 \). These can be factored further using the sum and difference of cubes formulas.
04

Apply the Difference of Cubes Formula

The formula for the difference of cubes is \( x^3 - y^3 = (x-y)(x^2+xy+y^2) \). Therefore, \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
05

Apply the Sum of Cubes Formula

The formula for the sum of cubes is \( x^3 + y^3 = (x+y)(x^2-xy+y^2) \). Therefore, \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).
06

Write the Complete Factored Form

Combine the factors obtained to write the complete factored form of the original polynomial: \( a^6 - b^6 = (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The **difference of squares** is a fundamental concept in algebra that involves factoring expressions where two squares are subtracted. Let's start with the general formula, which is \( x^2 - y^2 = (x - y)(x + y) \). This can be used when you have two perfect squares separated by a minus sign. The given polynomial \( a^6 - b^6 \) fits this pattern. We can rewrite it as \((a^3)^2 - (b^3)^2\), recognizing that both \(a^3\) and \(b^3\) are perfect squares.

  • The first step is to make each term a perfect square, which allows us to apply the difference of squares formula.
  • Then, we factor it into two binomials: \((a^3 - b^3)(a^3 + b^3)\).
This creates an opportunity to further factor each resulting binomial using the concepts of difference and sum of cubes.
Difference of Cubes
Once we've factored our polynomial using the difference of squares, we are left with two terms: \(a^3 - b^3\) and \(a^3 + b^3\). The **difference of cubes** is a specific way to factor a polynomial of the form \( x^3 - y^3 \), and the formula to use is \( x^3 - y^3 = (x-y)(x^2+xy+y^2) \).

  • In our exercise, applying this to \( a^3 - b^3 \), we get \( (a - b)(a^2 + ab + b^2) \).
  • This formula helps break down more complex polynomials and is useful when dealing with cubic terms.
Understanding this concept allows us to simplify the polynomial to a more manageable expression, ensuring every factor is in its most simplified form.
Sum of Cubes
The **sum of cubes** formula is employed when we have an expression in the form of \( x^3 + y^3 \). It might seem slightly more difficult, but it's straightforward with the right formula: \( x^3 + y^3 = (x+y)(x^2-xy+y^2) \). This allows us to factor sums of cubes.

  • Using this formula for \( a^3 + b^3 \) gives us \( (a + b)(a^2 - ab + b^2) \).
  • Just like the difference of cubes, the sum of cubes simplifies expressions that at first seem not easily factored.
Combining these techniques, we arrive at a fully factored expression for the original polynomial \( a^6 - b^6 \): \( (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) \). Understanding these processes makes complex polynomials much easier to handle.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Replace the symbol \(\square\) with either \(=\) or \(\neq\) to make the resulting statement true, whenever the expression has meaning. Give a reason for your answer. $$ \left(a^{\prime}\right)^{2} \square a^{(r)} $$

\(A\) cardboard box with an open top and a square bottom is to have a volume of \(25 \mathrm{ft}^{3}\). Use a table utility to determine the dimensions of the box to the nearest 0.1 foot that will minimize the amount of cardboard used to construct the box.

Choose the equation that best describes the table of data. (Hint: Make assignments to \(\mathbf{Y}_{\mathbf{r}}-\mathbf{Y}_{\mathbf{4}}\) and examine a table of their values.) $$\begin{array}{|c|c|}\hline x & y \\\\\hline 1 & 0.8 \\\\\hline 2 & -0.4 \\\\\hline 3 & -1.6 \\\\\hline 4 & -2.8 \\\\\hline 5 & -4.0 \\\\\hline\end{array}$$ (1) \(y=-1.2 x+2\) (2) \(y=-1.2 x^{2}+2\) (3) \(y=0.8 \sqrt{x}\) (4) \(y=x^{3 / 4}-0.2\)

Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt[3]{3 t^{4} v^{2}} \sqrt[3]{-9 t^{-1} v^{4}}$$

Apartment rentals \(A\) real estate company owns 218 efficiency apartments, which are fully occupied when the rent is 940 dollars per month. The company estimates that for each 25 dollars increase in rent, 5 apartments will become unoccupied. What rent should be charged in order to pay the monthly bills, which total 205,920 dollars ?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.