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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. 228 (page 604)

In the following exercises, factor completely.
${(y - 3)}^{3} - 64 y^{3}$

Short Answer

Expert verified

The simplified expression is $- (7 y + 3) (73 y^{2} - 30 y + 9)$

Step by step solution

01

Step 1. Given Information

The given expression is ${(y - 3)}^{3} - 64 y^{3}$

02

Step 2. Calculation

We will find the factor of binomial sum by using difference of cube pattern,

${(y - 3)}^{3} - 64 y^{3} = {(y - 3)}^{3} - {(8 y)}^{3} = ((y - 3) - 8 y) ({(y - 3)}^{2} + (y - 3) 8 y + {(8 y)}^{2}) = (y - 3 - 8 y) ({(y - 3)}^{2} + (8 y^{2} - 24 y) + 64 y^{2}) = (- 7 y - 3) ((y^{2} - 2 脳 y 脳 3 + 9) + (8 y^{2} - 24 y) + 64 y^{2}) = (- 7 y - 3) (73 y^{2} - 6 y - 24 y + 9) = - (7 y + 3) (73 y^{2} - 30 y + 9)$

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Most popular questions from this chapter

Factor $x^{2} - 7 x y - 10 y^{2}$

Factor using the 鈥榓c鈥� method:

$4 y^{2} + 8 y + 3$

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

$12 x^{3} - 10 x$

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

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

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

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