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(b) expansion for Hence (or otherwise) write down the form of the partial fraction x6 + x³ + 1 х8 — 1 without trying to solve for the resulting 8 constants. (First treat the denominator as a difference of two squares.)

(b) expansion for Hence (or otherwise) write down the form of the partial fraction x6 + x³ + 1 х8 — 1 without trying to solve for the resulting 8 constants. (First treat the denominator as a difference of two squares.)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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(a) Show that (x2 + v2x + 1)(x² – V2x + 1) = x4 + 1. | (b) expansion for Hence (or otherwise) write down the form of the partial fraction x6 + x³ + 1 x8 – 1 | without trying to solve for the resulting 8 constants. (First treat the denominator as a difference of two squares.)
Transcribed Image Text:(a) Show that (x2 + v2x + 1)(x² – V2x + 1) = x4 + 1. | (b) expansion for Hence (or otherwise) write down the form of the partial fraction x6 + x³ + 1 x8 – 1 | without trying to solve for the resulting 8 constants. (First treat the denominator as a difference of two squares.)
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Step 1

To write the form of the partial fraction expansion for:

x6+x3+1x8-1

 

 

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