College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.4 - Page 61: 55

Answer

$x^3-9x^2+27x-27$

Work Step by Step

$$(x-3)^3$$ $$=(x-3)(x-3)(x-3)$$ $$=(x-3)[(x\times x)+(x\times (-3))+(-3\times x)+(-3\times (-3))]$$ $$=(x-3)(x^2-3x-3x+9)$$ $$=(x-3)(x^2-6x+9)$$ $$=(x\times x^2)+(x\times (-6x))+(x\times 9)+(-3\times x^2)+(-3\times (-6x))+(-3\times 9)$$ $$=x^3-6x^2+9x-3x^2+18x-27$$ $$=x^3-9x^2+27x-27$$
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