College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 10

Answer

$(\sqrt{a}+\sqrt{b})^{6}=a^{3}+6a^{2}\sqrt{ab}+15a^{2}b+20ab\sqrt{ab}+15ab^{2}+6b^{2}\sqrt{ab}+b^{3}$ Or without roots: $(\sqrt{a}+\sqrt{b})^{6}=a^{3}+6a^{5/2}b^{1/2}+15a^{2}b+20a^{3/2}b^{3/2}+15ab^{2}+6a^{1/2}b^{5/2}+b^{3}$

Work Step by Step

We use the values from the 6th row of Pascal's Triangle to find the coefficients: $(\sqrt{a}+\sqrt{b})^{6}=1(\sqrt{a})^6+6(\sqrt{a})^5(\sqrt{b})^1+15(\sqrt{a})^4(\sqrt{b})^2+20(\sqrt{a})^3(\sqrt{b})^3+15(\sqrt{a})^2(\sqrt{b})^4+6(\sqrt{a})^1(\sqrt{b})^5+1(\sqrt{b})^6 =a^{3}+6a^{2}\sqrt{a}\sqrt{b}+15a^{2}b+20a\sqrt{a}b\sqrt{b}+15ab^{2}+6\sqrt{a}b^{2}\sqrt{b}+b^{3} =a^{3}+6a^{2}\sqrt{ab}+15a^{2}b+20ab\sqrt{ab}+15ab^{2}+6b^{2}\sqrt{ab}+b^{3}$ Or without roots: $a^{3}+6a^{5/2}b^{1/2}+15a^{2}b+20a^{3/2}b^{3/2}+15ab^{2}+6a^{1/2}b^{5/2}+b^{3}$
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