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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 6: Q. 6.63 (page 600)

Factor: $8 x^{3} - 27 y^{3}$ .

Short Answer

Expert verified

Factors of $8 x^{3} - 27 y^{3}$ are $(2 x - 3 y) (4 x^{2} - 6 x y + 9 y^{2})$ .

Step by step solution

01

Step 1. Given information.

Given an expression $8 x^{3} - 27 y^{3}$ .

02

Step 2. Check if the binomial is a difference and write the terms as cubes.

Consider $a = 2 x$ and $b = 3 y$ .

It can be seen that $8 x^{3} - 27 y^{3} = {(2 x)}^{3} - {(3 y)}^{3}$ .

Therefore, $8 x^{3} - 27 y^{3}$ is a difference of cubes.

03

Step 3. Use the difference of cubes pattern and simplify.

Note that $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ .

Factor the expression as follows.

$\begin{array}{rcl} 8 x^{3} - 27 y^{3} & = & {(2 x)}^{3} - {(3 y)}^{3} \\ = & (2 x - 3 y) ({(2 x)}^{2} + (2 x) (3 y) + {(3 y)}^{2}) \\ = & (2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2}) \end{array}$

04

Step 4. Check by multiplying the factors.

Verify as follows.

$\begin{array}{rcl} (2 x - 3 y) (4 x^{2} - 6 x y + 9 y^{2}) & = & 2 x (4 x^{2} - 6 x y + 9 y^{2}) - 3 y (4 x^{2} - 6 x y + 9 y^{2}) \\ = & 8 x^{3} - 12 x^{2} y + 18 x y^{2} - 12 x^{2} y + 18 x y^{2} - 27 y^{3} \\ = & 8 x^{3} - 27 y^{3} \end{array}$

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Most popular questions from this chapter

Factor by Grouping
In the following exercises, factor by grouping.

$4 x^{2} 鈭� 36 x 鈭� 3 x + 27$ .

In the following exercises, factor the greatest common factor from each polynomial.

$8 y^{3} + 16 y^{2}$

Factor completely using trial and error: $10 y^{2} - 37 y + 7$

Factor: $3 p^{3} - 6 p^{2} q + 9 p q^{3}$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c$ .

$q^{2} - 13 q + 36$

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