Example 6: Consider (x³ −7) € Q[x]. Let ae C be a root of (x³ -7) (in C). Find (1+ a)-¹. Q[x] Solution: {a+bx+cx²/a,b,ce Q1. request explain this step? Now a³ = 7. Also, since (1+x,x³ −7) = 1, by the Euclidean algorithm 3 f(x), g(x) = Q[x] such that (1+x)f(x) + (x³-7)g(x)=1, ~ In fact, here you can explicitly find f(x)=(x²- Now, putting x = α in (3), we get (1+x)f(x) = 1 )=1/(a²- i.e., (1+α)¹ = f(a)==(a²-a +1). 8 x²-x+1 xH) 2²-7 43 213 +x² -x2. -712 7 ан Quotient =x²-xH Remainder = -8 _n (x²-x+1) and g(x) n. 717 ...(3) How do we get this step o b(x) = f (x²-xH) 4. g(x) = ~{?
Example 6: Consider (x³ −7) € Q[x]. Let ae C be a root of (x³ -7) (in C). Find (1+ a)-¹. Q[x] Solution: {a+bx+cx²/a,b,ce Q1. request explain this step? Now a³ = 7. Also, since (1+x,x³ −7) = 1, by the Euclidean algorithm 3 f(x), g(x) = Q[x] such that (1+x)f(x) + (x³-7)g(x)=1, ~ In fact, here you can explicitly find f(x)=(x²- Now, putting x = α in (3), we get (1+x)f(x) = 1 )=1/(a²- i.e., (1+α)¹ = f(a)==(a²-a +1). 8 x²-x+1 xH) 2²-7 43 213 +x² -x2. -712 7 ан Quotient =x²-xH Remainder = -8 _n (x²-x+1) and g(x) n. 717 ...(3) How do we get this step o b(x) = f (x²-xH) 4. g(x) = ~{?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![3
Example 6: Consider (x³ -7)= Q[x]. Let ae C be a root of (x³-7) (in C).
Find (1+α)-¹.
Solution: €x]/<x²-7>* {a+bx+cx²| a, b,ce Q1. request explain this step?
Q[x]
→
<X
3
Now a³ = 7. Also, since (1+x, x³ − 7) = 1, by the Euclidean algorithm
X
3f(x), g(x) = Q[x] such that
(1+x)f(x) + (x³-7)g(x) = 1,
In fact, here you can explicitly find f(x) = (x² −x +1) and g(x) =
15
Now, putting x = α in (3), we get
(1+α)f(x)=1
i.e., (1+α)¯¹ = f(a) = — (a² − a +1).
8
Z-x+
H
xH) 22³-7
재)
213 +7²
2
-1²-7
2
-x
+-2
n --
Quotient
кан
Remainder = -8
구
Il t
...(3)
How do we get this step o
==}}
f(x) = {(x²-xH) 4 g(x) =
дса)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Fb330857d-df55-4dff-8077-06f2aac49deb%2Fqxvmew_processed.png&w=3840&q=75)
Transcribed Image Text:3
Example 6: Consider (x³ -7)= Q[x]. Let ae C be a root of (x³-7) (in C).
Find (1+α)-¹.
Solution: €x]/<x²-7>* {a+bx+cx²| a, b,ce Q1. request explain this step?
Q[x]
→
<X
3
Now a³ = 7. Also, since (1+x, x³ − 7) = 1, by the Euclidean algorithm
X
3f(x), g(x) = Q[x] such that
(1+x)f(x) + (x³-7)g(x) = 1,
In fact, here you can explicitly find f(x) = (x² −x +1) and g(x) =
15
Now, putting x = α in (3), we get
(1+α)f(x)=1
i.e., (1+α)¯¹ = f(a) = — (a² − a +1).
8
Z-x+
H
xH) 22³-7
재)
213 +7²
2
-1²-7
2
-x
+-2
n --
Quotient
кан
Remainder = -8
구
Il t
...(3)
How do we get this step o
==}}
f(x) = {(x²-xH) 4 g(x) =
дса)
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Request explain last step as to how do we get g(x)= -1/8 from the remainder and f(x)=-(1/8)*(x^2-x+1)
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