17 days without a single geometry problem. It's not that I don't like algebra but I love geometry and I enjoyed the geometric problems you post on your channel. I wish to see them back again quickly. Anis from Algeria.
I solved it in 5 sec It is directly a formula see We know:- (a+b) (b+c) (c+a) = a²b+ab²+b²c+bc²+a²c+ac²+2abc If we add abc to rhs it will be a²b+ab²+b²c+bc²+a²c+ac²+3abc which is (a+b+c) (ab+bc+ca)
Great video. Note that Vieta’s formulas involve the elementary symmetric polynomials, an interesting subject in itself. Also note that your identity shows up in (x+y+z)³ = x³+y³+z³+3(x+y)(x+z)(y+z).
I saw on your Twitter that you are into Cybersecurity Ironically as I'm typing this, I'm in Cybersecurity university course this video seems interesting :D
I have a somewhat nasty solution that requires no finesse. Let it = 0 and use the quadratic formula to solve for x. There are two roots: x=-(y+z) and x=-yz/(y+z). From there you have the factors.🤣
@@SyberMath I prefer the first method. It is easier to understand, even if it takes longer. The only reason I say that is because I am not familiar with the Vieta Way.
I distribute the terms also but i rewrite the sum as a quadratic with x as the variable (y+z)x^2 + ( y^2 + 3yz + z^2 )x + yz(y+z) => (y+z)x^2 + (y+ z)^2*x + yz*x + yz(y+z) => (y+z)x*( x + (y+z) ) + yz*( x + (y+z) ) => ( x + (y+z) )*( (y+z)x + yz ) => (x+y+z)(xy+yz+xz) Also great video
Amazing problem !!!!!! [(x+y+z)-z][(x+y+z)-y][(x+y+z)-x] + xyz Finaly you get -xyz and all the rest term have (x+y+z) as factor so we get finaly (x+y+z)(xy+xz+yz) and yay!!!🎉🎊🎉🎊
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
I enjoy the way you manipulate these symmetric polynomials with such facility!
Glad you like them!
The first method is what I thought of initially and is pretty straightforward. The second solution is unrealistically concise! Well done once again!
Thank you!
17 days without a single geometry problem. It's not that I don't like algebra but I love geometry and I enjoyed the geometric problems you post on your channel. I wish to see them back again quickly. Anis from Algeria.
Sure. I understand. I will do a geometry puzzle after tomorrow. and you're the first person to know that!
🙂
@@SyberMath nahh i saw it before him :D am the first xD
@@dhruvladdha4789 Oh ok! 😁
The first way was straightforward and easier for me to follow. Love algebra.
Good to hear!
I solved it in 5 sec
It is directly a formula see
We know:-
(a+b) (b+c) (c+a) = a²b+ab²+b²c+bc²+a²c+ac²+2abc
If we add abc to rhs it will be
a²b+ab²+b²c+bc²+a²c+ac²+3abc which is
(a+b+c) (ab+bc+ca)
Great video. Note that Vieta’s formulas involve the elementary symmetric polynomials, an interesting subject in itself.
Also note that your identity shows up in (x+y+z)³ = x³+y³+z³+3(x+y)(x+z)(y+z).
Good point!
Put x+y+z = 0 we get function as zero. Means x+y+z is a factor. By symmetry and inspection next bracket can be guessed.
That's right!
I saw on your Twitter that you are into Cybersecurity
Ironically as I'm typing this, I'm in Cybersecurity university course
this video seems interesting :D
That's interesting! Do we follow each other?
@@SyberMath Hey I just made a twitter account and followed you. I tweeted you
Can you follow me back?
I used exactly the first method, but the second one is interesting as well. I knew it has something to do with Vieta's formulas. Nicely done.
Thank you, Snejpu!
Hey, I've learnt something new from your video and it is the method no. 2.It was really very interesting
Glad to hear that!
Take x=-y-z, we have (-z)*(-y)*(y+z)+(-y-z)yz=0. So, the formula has the factor x+y+z. Then we can use method 2 in progress.
Nice!
I have a somewhat nasty solution that requires no finesse. Let it = 0 and use the quadratic formula to solve for x. There are two roots: x=-(y+z) and x=-yz/(y+z). From there you have the factors.🤣
That's interesting!
Another great explanation!
Thank you!
wow that second method was AMAZING Everything just cancelled with it's additive inverse
Yeah, I like the second method because it's the Vieta way!
@@SyberMath I prefer the first method. It is easier to understand, even if it takes longer. The only reason I say that is because I am not familiar with the Vieta Way.
I distribute the terms also but i rewrite the sum as a quadratic with x as the variable
(y+z)x^2 + ( y^2 + 3yz + z^2 )x + yz(y+z) => (y+z)x^2 + (y+ z)^2*x + yz*x + yz(y+z) =>
(y+z)x*( x + (y+z) ) + yz*( x + (y+z) ) => ( x + (y+z) )*( (y+z)x + yz ) =>
(x+y+z)(xy+yz+xz)
Also great video
Second method was the one which I used to factor this equation by seeing in the thumbnail. Just using a instead of b
Nice!
Excellent vid. I liked the second method👍
Glad you liked it!
Second method just shows the beauty of Algebra MIND BLOWN
🤩🤩🤩
Nice one
Thanks 🔥
"Can you find the factor(s) or try to factorize", should be the question..
alternative: factorise (in the UK)
'factor' can be used as a verb as well as a noun!
Ça y est, j’ai trouvé !!!!
très bien!!!
nice question
Thank you!
seems easy but requires a bit of thinking
HELLO ! KAISE HO!
Yes! Thank you!
Hi!
Amazing problem !!!!!!
[(x+y+z)-z][(x+y+z)-y][(x+y+z)-x] + xyz
Finaly you get -xyz and all the rest term have (x+y+z) as factor so we get finaly (x+y+z)(xy+xz+yz) and yay!!!🎉🎊🎉🎊
Nice!!! 😊