Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises: 45

Answer

$ 1 - 6r + 12r^2 - 8r^3$

Work Step by Step

$Multiply$ $the$ $algebraic$ $expressions$ $using$ $a$ $Special$ $Product$ $Formula$ $and$ $simplify:$ $(1-2r)^3$ Use the Cube of a Difference Formula: $(A-B)^3 = A^3 - 3A^2B + 3AB^2 - B^3$ $(1-2r)^3$ = $1^3$ - $(3\times 1^2 \times 2r)$ + $(3\times 1 \times (2r)^2)$ - $(2r)^3$ $= 1 - 6r + 12r^2 - 8r^3$
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