Answer
$j^{3}+9j^{2}k+27jk^{2}+27k^{3}$
Work Step by Step
Expand $(j+3k)^{3}$ using Pascal's Triangle:
$(j+3k)^{3}=j^{3}+3j^{2}(3k)+3j(3k)^{2}+(3k)^{3}$.
Multiply the terms:
$j^{3}+9j^{2}k+27jk^{2}+27k^{3}$.
Algebra 2 Common Core
by Hall, Prentice
Published by
Prentice Hall
ISBN 10:
0133186024
ISBN 13:
978-0-13318-602-4
Chapter 11 - Probability and Statistics - Get Ready! - Page 671: 9
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