Explanation The given equation is, n 2 − 1 n 3 + 1 = A n 2 − n + 1 . Reduce the numerator and denominator of the fraction on the left hand side of the equation as follows. ( n + 1 ) ( n − 1 ) ( n + 1 ) ( n 2 − n + 1 ) = A n 2 − n + 1 ( ∵ a 2 − b 2 = ( a + b ) ( a − b ) and a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) ) Cancel the like terms in each denominator of the equation in order to find the value of A . ( n + 1 ) ( n − 1 ) ( n + 1 ) = A 1 A = ( n + 1 ) ( n − 1 ) ( n + 1 ) ( Cancel like terms ) A = n − 1 Thus, the value of A is n − 1

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 6.4, Problem 24E

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The value of A such that the fractions in n2−1n3+1=An2−n+1 are equivalent.

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Chapter 6.4, Problem 23E

Chapter 6.4, Problem 25E

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Chapter 6 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 6.1 - Prob. 1PE Ch. 6.1 - Prob. 2PE Ch. 6.1 - Prob. 3PE Ch. 6.1 - Prob. 4PE Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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The value of A such that the fractions in n 2 − 1 n 3 + 1 = A n 2 − n + 1 are equivalent.

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The value of A such that the fractions in n2−1n3+1=An2−n+1 are equivalent.

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Chapter 6 Solutions