Answer
$x=2$ with a multiplicity of $3$.
Work Step by Step
$x^3-8=0$ (set equal to zero, and add $8$ to the other side)
$x^3=8$ (take the cube root of 8).
$\sqrt[3] 8 = 2$
Since there is a degree of $3$, your answer has a multiplicity of $3$.
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 293: 16
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