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Chapter 7: Problem 78

Factor completely. $$n^{3}+125$$

Short Answer

Expert verified
The completely factored form of \(n^3 + 125\) is \((n + 5)(n^2 - 5n + 25)\).

Step by step solution

01

Identify the sum of cubes

Given the expression \(n^3 + 125\), we can rewrite it as a sum of two cubes: \(n^3 + 5^3\). Here, \(a = n\) and \(b = 5\).
02

Apply the sum of cubes formula

Now, apply the sum of cubes formula \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\) to our expression: \((n^3 + 5^3) = (n + 5)(n^2 - n\cdot5 + 5^2)\)
03

Simplify the expression

Finally, we simplify the expression: \((n^3 + 5^3) = (n + 5)(n^2 - 5n + 25)\) So, the completely factored form of the given expression is \((n + 5)(n^2 - 5n + 25)\).

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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
The sum of cubes is a concept used in algebra to simplify expressions that are in the form of adding two cubes together. Imagine you have two numbers, say \(a\) and \(b\), and you want to add their cubes. This is represented as \(a^3 + b^3\).
To factor the sum of cubes, we use the formula:
  • \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\)
This formula helps break down a seemingly complex expression into simpler components. For example, in the original exercise of \(n^3 + 125\), we can identify it as a sum of cubes, where \(a = n\) and \(b = 5\). Applying the formula allows us to factor it easily.
Factoring Polynomials
Factoring polynomials is all about breaking down a complex polynomial expression into simpler elements or factors that are multiplied together to give the original polynomial. It involves identifying any common factors first and then applying various factoring techniques.
For example, when you encounter the polynomial \(n^3 + 125\), you can look for special patterns like the sum of cubes or difference of cubes. Recognizing the sum of cubes pattern allows us to apply the appropriate formula which makes factoring straightforward.
This skill is crucial because it simplifies expressions, making further mathematical operations easier to handle.
Breaking Down Algebraic Expressions
Algebraic expressions are made up of terms that include numbers, variables, and arithmetic operations. The goal in breaking them down is to make complex expressions simpler to work with.
  • In expressions like \(n^3 + 125\), the task is to identify patterns such as cubes, quadratics, or other recognizable forms so we can use specific formulas to simplify them.
  • This process often involves recognizing when to apply formulas like the sum of cubes, quadratic formulas, or factoring by grouping.
Understanding how to work with algebraic expressions allows for greater ease in solving equations, graphing functions, and performing further algebraic manipulations.

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Most popular questions from this chapter

Factor completely. $$216 a^{3}+64 b^{3}$$

Factor completely by first taking out a negative common factor. $$-16 y^{2}-34 y+15$$

Factor by grouping. $$40 j^{3}+72 j k-55 j^{2} k-99 k^{2}$$

Write an equation and solve. The longer leg of a right triangle is \(7 \mathrm{cm}\) more than the shorter leg. The length of the hypotenuse is \(3 \mathrm{cm}\) more than twice the length of the shorter leg. Find the length of the hypotenuse.

Factor completely. $$\frac{1}{4} k^{2}-\frac{4}{9}$$
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