Answer
The required solution is ${{\left( c+3 \right)}^{5}}={{c}^{5}}+15{{c}^{4}}+90{{c}^{3}}+270{{c}^{2}}+405c+243$
Work Step by Step
See below:
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1092: 24
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