I need a help with this problem
M MathBuff New member Joined Mar 11, 2020 Messages 4 Mar 11, 2020 #1 I need a help with this problem Attachments 6.PNG 10.8 KB · Views: 9
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Mar 11, 2020 #2 I would begin by observing that: [MATH]x^3-1=(x-1)(x^2+x+1)[/MATH] And so we know the roots of \(x^2+x+1\) are complex cube roots of unity. Can you proceed?
I would begin by observing that: [MATH]x^3-1=(x-1)(x^2+x+1)[/MATH] And so we know the roots of \(x^2+x+1\) are complex cube roots of unity. Can you proceed?
M MathBuff New member Joined Mar 11, 2020 Messages 4 Mar 11, 2020 #3 Can i get more help? Cause i still don't know, sorry
firemath Full Member Joined Oct 29, 2019 Messages 602 Mar 11, 2020 #4 @MathBuff, you need to at least try what MarkFL has given you. Don't give up yet!
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Mar 11, 2020 #5 MathBuff said: Can i get more help? Cause i still don't know, sorry Click to expand... Suppose \(\beta\) is a complex cube root of unity. Then we must have: [MATH]\beta^{3k}=(\beta^3)^k=1^k=1[/MATH] where \(k\in\mathbb{N}\). And since \(f(x)\) is divisible by \(x^2+x+1\) what must \(f(\beta)\) be? Hint: [MATH]f(x)=(x^2+x+1)P(x)[/MATH]
MathBuff said: Can i get more help? Cause i still don't know, sorry Click to expand... Suppose \(\beta\) is a complex cube root of unity. Then we must have: [MATH]\beta^{3k}=(\beta^3)^k=1^k=1[/MATH] where \(k\in\mathbb{N}\). And since \(f(x)\) is divisible by \(x^2+x+1\) what must \(f(\beta)\) be? Hint: [MATH]f(x)=(x^2+x+1)P(x)[/MATH]