For this to work, you would have to know ahead of time that a factor of (x^3 - 1) would be a factor of the initial expression. And how could you possibly know that in advance? I would like to see a video explaining why you made that seemingly arbitrary decision at the beginning.
@Chris Sekely I agree with Chris Sekely, Isn't this a "saw the answer and come up with a solution" type? Who in exam or competition will think of adding a x^10 ???
The fundamental idea is that cube roots of unity satisfy this property of x^2+x+1=0 The polynomial in the expression can be reduced to this form . If you do a little manipulation. Difficult for me to write this here
@@rajkumarbajagain5392 You should study the properties of complex roots of unity . They follow this x^2+x+1=0 If you study any book which treats complex numbers in a little detail. You will start realising these instantly
Well Done Young man. You are real Professor of Algebraic Techniques. You have all the skills to train even dummies. I solute you, I am always benefitted from your excellent skills. Thanks
Done this in my mind a lot quicker: x^13+x^11+1=x^11*(x^2+x+1)-(x^12-1)=x^11*(x^2+x+1)-(x-1)(x^11+x^10+...+1)=x^11*(x^2+x+1)-(x-1)*(x^2+x+1)(x^9+x^6+x^3+1) Watched the video hoping to see how the remaining 11th power polynomial is factored, but it's...not. Upd: can be done even easier literally in 2 seconds. In order to be divisible by x^2+x+1 a polynomial obviously has to have equal summs of coefficients of powers with the same power mod 3, i.e.: a0+a3+a6+...=a1+a4+a7+...=a2+a5+a8+..., which is easily verifiable in this case.
Why didn’t you go for one step and substructure x : x^4-x? You would get directly x^2+x+1 instead of x^4+x^2+1and the common factor with x^3-1 would show up immediately.
What about the rest (the 11th deg polynomial)? It looks factoring == to break it down into product of 1st and 2nd degree polynomials. This is factoring by grouping and looks very difficult to handle.
@@MizardXYT it's more general than that. Consider ω, a cube root of unity. So long as a and b are different non-zero residues mod 3, it will work (by factor theorem).
For this to work, you would have to know ahead of time that a factor of (x^3 - 1) would be a factor of the initial expression. And how could you possibly know that in advance? I would like to see a video explaining why you made that seemingly arbitrary decision at the beginning.
Good point, Chris! I'm planning to make a video explaining these types of expressions.
@Chris Sekely
I agree with Chris Sekely, Isn't this a "saw the answer and come up with a solution" type? Who in exam or competition will think of adding a x^10 ???
The fundamental idea is that cube roots of unity satisfy this property of x^2+x+1=0
The polynomial in the expression can be reduced to this form . If you do a little manipulation.
Difficult for me to write this here
ya I am also thinking how to get the idea that there is a factor of x^3-1?
@@rajkumarbajagain5392
You should study the properties of complex roots of unity .
They follow this x^2+x+1=0
If you study any book which treats complex numbers in a little detail.
You will start realising these instantly
Most equations of the form x^(n+2) + x^n + 1 = 0 and x^(n+1) + x^n + 1 = 0 can be reduced by considering x³ = 1
I like how the terms in the 2nd factor have alternating signs. Anyone else noticed that?
Well Done Young man. You are real Professor of Algebraic Techniques. You have all the skills to train even dummies. I solute you, I am always benefitted from your excellent skills. Thanks
Wow, thanks for the compliment! 🤩🧡
Done this in my mind a lot quicker: x^13+x^11+1=x^11*(x^2+x+1)-(x^12-1)=x^11*(x^2+x+1)-(x-1)(x^11+x^10+...+1)=x^11*(x^2+x+1)-(x-1)*(x^2+x+1)(x^9+x^6+x^3+1)
Watched the video hoping to see how the remaining 11th power polynomial is factored, but it's...not.
Upd: can be done even easier literally in 2 seconds. In order to be divisible by x^2+x+1 a polynomial obviously has to have equal summs of coefficients of powers with the same power mod 3, i.e.:
a0+a3+a6+...=a1+a4+a7+...=a2+a5+a8+..., which is easily verifiable in this case.
Happy 700 subs!!! YAY
Thanks to you and all the great people that keep watching!!!
@@SyberMath man you are bringing up awesome teasers ,you are actually great ,making days great for me
Frumos! Calcul bine aranjat si muncit! Next!?
Superb explanation 👍
Thank you 🙂
faster method : P(j)=j^13+j^11+1=j+j²+1=0
P(j²)=P(conj(j))=0 Factorise with (x-j)(x-j²)=x²+x+1
then divide P(x) by x²+x+1 to obtain other factor
6:52, why not extract x^2 at the same time?
Why didn’t you go for one step and substructure x : x^4-x? You would get directly x^2+x+1 instead of x^4+x^2+1and the common factor with x^3-1 would show up immediately.
That's right!
Nice!
What about the rest (the 11th deg polynomial)?
It looks factoring == to break it down into product of 1st and 2nd degree polynomials.
This is factoring by grouping and looks very difficult to handle.
This is a tough polynomial!
Yes indeed!
Problem fixed. Re-recorded:
ua-cam.com/video/TYqmZRo0gCc/v-deo.html
@@SyberMath Great. Thanks!
eek when i saw this i so much freaked out it looks gigantic and scary! but you are so cool you froze it.
Thank you!!! 🧡
This is a clever factorization problem.
Thank you!!!
Is this a "saw the answer and come up with a solution" type? Who in exam or competition will think of adding a x^10 ???
There's a reason why this works
Good one
Thanks
I swear to god, x^a + x^b + 1 HAS to be divisble by x^2+x+1. Happened too many times.
There's a secret behind it! I'll explain that later! 😁
x^(3*n+1) + x^(3*n-1) + 1 = (x² + x + 1)*(x^(3*n-1) + (1 - x)*(sum(x^(3*k), k = 0..n-1)))
n = 4 gives
x¹³ + x¹¹ + 1 = (x² + x + 1)*(x¹¹ + (1 - x)*(1+x³+x⁶+x⁹)) = (x² + x + 1)*(x¹¹-x¹⁰+x⁹-x⁷+x⁶-x⁴+x³-x+1)
@@MizardXYT it's more general than that. Consider ω, a cube root of unity. So long as a and b are different non-zero residues mod 3, it will work (by factor theorem).
@@SyberMath Did you do this yet? I haven't spotted it.
@@pietergeerkens6324 Not yet. Thanks for the reminder!
So great.
Thanks
This video is a great example of "Just because you can, doesn't mean you should".
Exactly
Excellent!
Many thanks!
Just apply x11 =a
Then a2+ a+1 = 0
x muliplied by xto the power 8 is equal to x to the power 8, well done : then u correct is showing you in poor light
Did I say that?
more scareful than " Shining " movie ... 🙃
You solved it but did not explain why you did what you did.
Sorry about that
feels like cheating hehe:
*Factor[x^13 + x^11 + 1]*
Yessssss! 😠😳😂
👍
you are insane
That's so true! 😲😂