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Chapter 6: Problem 27

Factor the given expressions completely. $$64-x^{6}$$

Short Answer

Expert verified
\((2-x)(4+2x+x^2)(2+x)(4-2x+x^2)\).

Step by step solution

01

Identify the Form of the Expression

The expression given is \(64 - x^6\). This suggests a difference of squares, since \(64\) is a perfect square \((8)^2\) and \(x^6\) can be written as \((x^3)^2\). Hence, the expression is a difference of squares \(a^2 - b^2\), where \(a = 8\) and \(b = x^3\).
02

Apply the Difference of Squares Formula

The difference of squares formula states: \(a^2 - b^2 = (a + b)(a - b)\). Applying this formula to our expression: \(64 - x^6 = (8 + x^3)(8 - x^3)\).
03

Check for Further Factoring

Look at the factors \(8 + x^3\) and \(8 - x^3\) to see if they can be factored further. The expression \(8 - x^3\) is a difference of cubes, and \(8 + x^3\) is a sum of cubes.
04

Factor the Difference of Cubes

Use the formula for the difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). For \(8 - x^3\), let \(a = 2\) and \(b = x\). This gives: \[8 - x^3 = (2-x)(4+2x+x^2).\]
05

Factor the Sum of Cubes

Use the formula for the sum of cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). For \(8 + x^3\), let \(a = 2\) and \(b = x\). This gives: \[8 + x^3 = (2+x)(4-2x+x^2).\]
06

Write the Completely Factored Expression

Combine the factored terms together. The complete factorization of \(64 - x^6\) is: \[(2-x)(4+2x+x^2)(2+x)(4-2x+x^2).\]

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Key Concepts

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

Difference of Squares
The concept of the difference of squares is a foundational topic in algebra. It involves expressions in the form of \(a^2 - b^2\), where both \(a\) and \(b\) are squared terms. The beauty of this concept lies in its factorization simplicity using the formula:
  • \(a^2 - b^2 = (a + b)(a - b)\)
Breaking it down: you identify \(a\) and \(b\) such that \(a = \sqrt{\text{first term}}\) and \(b = \sqrt{\text{second term}}\). For example, for the expression \(64 - x^6\), we rewrite it as \((8)^2 - (x^3)^2\). By applying the formula, we achieve two factors: \((8 + x^3)\) and \((8 - x^3)\). This step not only simplifies the expression but also prepares it for further factorization, if possible.
Difference of Cubes
When confronted with an expression that's a difference of cubes, the factorization takes a unique yet fascinating approach. The formula for the difference of cubes is:
  • \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
In simpler terms, \(a\) and \(b\) are cube roots of the individual terms. Let's take \(8 - x^3\) for example. Recognizing that \(a = 2\) and \(b = x\) (since \(8 = 2^3\) and \(x^3 = x^3\)), we apply the formula:
  • \((2 - x)(4 + 2x + x^2)\)
This provides a complete breakdown, transforming the complex cube difference into simpler factors. This type of simplification is tremendously useful as it reduces equations into manageable parts.
Sum of Cubes
The sum of cubes might seem challenging, but it also has a straightforward formula like the difference of cubes. The formulas help break down the problem into smaller, workable parts:
  • \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
Using the sum of cubes formula, dealing with terms like \(8 + x^3\) becomes effortless. Here, \(8 = 2^3\) and \(x^3 = x^3\) suggest \(a = 2\) and \(b = x\). By applying the formula, we derive the factors:
  • \((2 + x)(4 - 2x + x^2)\)
Understanding this process facilitates deeper algebraic factorization and reveals simplicity in seemingly complex expressions.
Perfect Squares
Perfect squares emerge when a number or expression is multiplied by itself. Recognizing perfect squares is crucial, especially when identifying factorization patterns like the difference or sum of squares.
  • Common examples include expressions like \(a^2\), where \(a\) is any real number or variable
  • 64 is a perfect square since \(8^2 = 64\)
In algebra, perfectly squared numbers form beautiful patterns that simplify the more complicated-looking expressions. Another example is \(x^6\). It can be seen as \((x^3)^2\), making it a perfect square. This recognition aids in identifying formulas or shortcuts that can further simplify expressions, like transforming them into a difference or sum of squares for easier factorization. By honing these skills, tackling polynomial challenges becomes more intuitive.

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Most popular questions from this chapter

Set up appropriate equations and solve the given stated problems. All numbers are accurate to at least two significant digits. A total of \(276 \mathrm{m}^{3}\) of water was pumped from a basement for \(4 \mathrm{h}\) 50 min. The pumping rate for part of the time was \(60.0 \mathrm{m}^{3} / \mathrm{h}\), and was then reduced to \(18 \mathrm{m}^{3} / \mathrm{h}\). What volume was pumped at the lower rate?

Solve the given equations and check the results. $$\frac{x}{5}+2=\frac{15+x}{10}$$

Solve for the indicated letter. Each of the given formulas arises in the technical or scientific area of study listed. \(P=\frac{R T}{V-b}-\frac{a}{V^{2}},\) for \(T \quad\) (thermodynamics)

Perform the indicated operations. Each expression occurs in the indicated area of application. $$\frac{\frac{L}{C}+\frac{R}{s C}}{s L+R+\frac{1}{s C}} \text { (electronics) }$$

Set up appropriate equations and solve the given stated problems. All numbers are accurate to at least two significant digits. A jet takes the same time to travel \(2580 \mathrm{km}\) with the wind as it does to travel \(1800 \mathrm{km}\) against the wind. If its speed relative to the air is \(450 \mathrm{km} / \mathrm{h},\) what is the speed of the wind?
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