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: 57

Answer

$27x^3-108x^2+144x-64$

Work Step by Step

$$(3x-4)^3$$ $$=(3x-4)(3x-4)(3x-4)$$ $$=(3x-4)[(3x\times 3x)+(3x\times (-4))+(-4\times 3x)+(-4\times (-4))]$$ $$=(3x-4)(9x^2-12x-12x+16)$$ $$=(3x-4)(9x^2-24x+16)$$ $$=(3x\times 9x^2)+(3x\times (-24x))+(3x\times 16)+(-4\times 9x^2)+(-4\times (-24x))+(-4\times 16)$$ $$=27x^3-72x^2+48x-36x^2+96x-64$$ $$=27x^3-108x^2+144x-64$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.