A modified method is to carry out the indicated multiplication in the numerator, then split the result into the two pieces x^7+y^7+z^7 and x^2*y^5+x^5*y^2+z^2*(x^5+y^5)+z^5*(x^2+y^2). The first piece divided by the denominator gets you 1. After substituting -x-y for z in the second part and doing the same in the denominator, arithmetic gets you the additional 3/7 but without the complications of difficult factoring.
Another great vid, Syber! Of course I didn't even think about attempting to solve this one, since it's way too much work to fit on my small whiteboard, but it was fun watching you walk us thru the steps.
Wow thats so cool, it's just like following your instincts that tell you to multiply the powers in the numerator (which gives 10) , divide it by the power in the denominator (which is 7), and get 10/7 😊
You could have assumed x, y, z as roots of a cubic polynomial. Then sum of roots taken one at a time becomes 0, and then use Newton's sum, which simplifies easily as a recurrence relation.
depending on what one wants to see as result: *Reduce[{x + y + z == 0, ((x^2 + y^2 + z^2) (x^5 + y^5 + z^5))/(x^7 + y^7 + z^7) == a, x > 0, y > 0}, {a, x, y, z}] // TraditionalForm*
As u told in 11:32 that this can be done in another video I would like to share my idea which is I did same thing in 1 taking 2 as common expression but in other two I just took 7 and 5 as a factor nothing else and then I multiply 1 ex. and 2ex. and found that in the last expression taking 7 as common looks the same then those got cancel and end up with 10/7.
I'm not sure how you would prove it, but Given x+y+z=0 and (Sum from i=0 to i=n (a_i)) =(Sum from j=0 to j=m (b_j)) where each a_i and b_i is a prime>1. Then (Product from i=0 to i=n (x^(a_i)+y^(a_i)+z^(a_i))) /(Product from j=0 to j=m(x^(b_j)+y^(b_j)+z^(b_j))) =(Product from i=0 to i=n (-1^ceiling(a_i/2)*a_i)) /(Product from j=0 to j=m(-1^ceiling(b_j/2)*b_j))
Dude! This problem was so complex, if I had tried this before the video I wud hav used up like 10 pages and realize none of the approach works, and it becomes so complex that my noob brain can't handle it
Observe that the given problem is equivalent to sum of roots of cubic eqation in t as 0 .Now use vietas method and recurrence to get it sorted out fast ! Anyway ur method was also nice
Your method is slick bro!!! If S[n]=x^n+y^n+z^n then the problem becomes S[2]*S[5]/S[7].... using vieta's and forming a recursion involving S[n] I had letters cancel out and I only bothered about S[2] which is -2(xy+yz+xz) or for me is -2b since b is the coefficient of x in the cubic equation....involving recursion, the rest was history
use Newton's identities for power sums and elementary symmetric polynomials. for 3 variables (the roots of some cubic): e1=p1=x+y+z=0 e2=xy+xz+yz e3=xyz p2=(e1)(p1)-2(e2) = -2(e2) for j>2: pj = (e1,-e2,e3) dot (pj-1,pj-2,pj-3) and using p1=e1=0: p3=3(e3) p4=2(e2)^2 p5=-5(e2)(e3) p6=-2(e2)^3 + 3(e3)^3 p7=7(e2)^2(e3) therefore: (p2)(p5)/(p7) = (-2(e2))(-5(e2)(e3))/[7(e2)^2(e3)] 10(e2)^2(e3)/[7(e2)^2(e3)] = 10/7 []
Now that was really tough to solve. I didn't try this out, because I thought I have to raise x+y+z to the 5th and 7th powers... That probably wouldn't help me at all, solving this problem required quite a lot of invention. Well, you definitely made up for this "one liner" from a few days ago, where result was -3 and the video could have lasted 1 minute. : ) Keep up the good content.
Thank you! The point for that video was how painful a solution can be if you do not follow some shortcuts or some elegant method. I like introducing two methods because everyone learns differently, it gives a chance to people to compare and contrast, and it's good for improving problem solving skills.
@@SyberMath Lovely. I appreciate it. I also appreciate that despite your channel is growing, you still read every comment under the videos. That's important.
Thank you for your efforts ... You have good videos and good presentation of problem solutions. I hope that your channel will continue to grow. Develop further and you will succeed. I wish you all the best ... See you soon ...
I’m these problems typically if you want to solve fast you can assume that plugging any number won’t make any difference ( unless you plug in a value that makes the denominator undefined ) Plug in -1 and -1 and 2 and you will get : 1+1+4=6 -1+-1+32=30 180/(-1+-1+128) =180/126=20/14=10/7 Done!
Maybe start with plotting to find its constant and then... Should just 1 point (or is it line or surface) omit, where x^7+y^7+z^7=0 ? (To avoid division by zero) The validity at division by zero could be possible proven with the limes testing. If for some reason the original equation is not important here this should be correct.
If (x^a+y^a+z^a)(x^b+y^b+z^b)/(x^c+y^c+z^c) can be simplified to a constant under the condition x+y+z=0, then the value must equal what you get if you take (x, y, z) = (1, -½, -½), which is one point satisfying the condition. The value then becomes (1+2(-½)^a)(1+2(-½)^b)/(1+2(-½)^c), which does not equal ab/c in the general case.
@@md2perpe you're probably right. It's easy to check it out but I'm not tempted cuz it doesn't look like there's an elegant solution for this problem (unless induction will do the trick): it was solved by brute force
Let xⁿ+yⁿ+zⁿ=P(n) (x+y+z)²=0, x²+y²+z²=-2(xy+xz+yz) (x+y+z)(x²+y²+z²-xy-xz-yz)=x³+y³+z³-3xyz=0, x³+y³+z³=3xyz Using the identity P(n)=P(n-1)(x+y+z)-P(n-2)(xy+xz+yz)+P(n-3)xyz=P(n-2)P(2)/2+P(n-3)P(3)/3 P(4)=P(2)²/2 P(5)=P(2)P(3)/2+P(2)P(3)/3=5/6(P(2)P(3)) P(7)=5/12(P(2)²P(3))+1/6(P(2)²P(3))=7/12(P(2)²P(3)) P(2)P(5)/P(7)=(5/6P(2)²P(3))/(7/12P(2)²P(3))=10/7 ab/c hypotesis doesn't work because if we substitute P(3)P(4)/P(7) yielding 6/7 instead of 6/7
@@SyberMath I guess that x+y+z=0 or x^n+y^n+z^n = 0 and [(x^{n+1}+y^{n+1}+z^{n+1})(x^{n+3}+y^{n+3}+z^{n+3})]/(x^{2n+4}+y^{2n+4}+z^{2n+4})=C I'll check with google sheet
Symetrial polynomials and basic theorem so the nominator and denominator can be written in polynomial of (x+y+z)^n + something. Due condition all is about what last number of linear term is in nominator and denominator and this is answer. When we see it better the answer is possible to see inmediatelly :) Too easy.
So how you can generalize this statement? For which n and m (x^n+y^n+z^n)*(x^m+y^m+z^m)/(x^(n+m)+y^(n+m)+z^(n+m)) is independent of the choice of x, y, z?
Using the results of another video you only need to find out what (x²+y²+z²)(x³+y³+z³) is, however I get -10/7. Either I made a mistake or you. Ok, found it. ;-)
I have a question: Are there infinitely many solutions for this equation? (15)^2a+15b+c=13 If there is a finite number of solutions: which are they? And if there are infinitely many solutions and I use other equations would there be less solutions? Please help me I need this because I want to make a graph in an unusual way... Thanks
A modified method is to carry out the indicated multiplication in the numerator, then split the result into the two pieces x^7+y^7+z^7 and x^2*y^5+x^5*y^2+z^2*(x^5+y^5)+z^5*(x^2+y^2). The first piece divided by the denominator gets you 1. After substituting -x-y for z in the second part and doing the same in the denominator, arithmetic gets you the additional 3/7 but without the complications of difficult factoring.
Another great vid, Syber! Of course I didn't even think about attempting to solve this one, since it's way too much work to fit on my small whiteboard, but it was fun watching you walk us thru the steps.
You can do it! You just need a bigger whiteboard 😁
When I also looked at x+y+z=0 with similar arrangements of (x^m+y^m+z^m)*(x^n+y^n+z^n)/(x^(m+n)+y^(m+n)+z^(m+n)) for m+n
Just amazing!...
Let's generalize (2, 5, 7) to (a, b, c). I wrote a script that looked at non-zero |a|, |b|, |c|
Wow thats so cool, it's just like following your instincts that tell you to multiply the powers in the numerator (which gives 10) , divide it by the power in the denominator (which is 7), and get 10/7 😊
Try it for (square)x(cube)/(5th power). Your approach would give 2x3/5 = 6/5. The answer is... (left as an exercise for the reader).
@@adandap 3,4,7 doesn't work
Took me 3 pages of work to do because I massively overcomplicated it. You did it in such a simple way and I’m actually fuming lol
Sorry! 😁
You could have assumed x, y, z as roots of a cubic polynomial. Then sum of roots taken one at a time becomes 0, and then use Newton's sum, which simplifies easily as a recurrence relation.
Your channel is growing fast! And I will keep supporting you! You are a pro!
I appreciate it!
Congrats on 19K wow
Really nice problem too
Thank you, Math Elite! Congrats on 2 • 3 • 157 subs! 😁
great job Syber, this was complicated but you solved it easily
Thank you! 😊
depending on what one wants to see as result:
*Reduce[{x + y + z == 0, ((x^2 + y^2 + z^2) (x^5 + y^5 + z^5))/(x^7 + y^7 + z^7) == a, x > 0, y > 0}, {a, x, y, z}] // TraditionalForm*
Mathematica software in action :)
@@vladimirzahoransky4505 the only way i can do maths :D
this is so brilliant!!!
Thank you!!! 💖
As u told in 11:32 that this can be done in another video I would like to share my idea which is I did same thing in 1 taking 2 as common expression but in other two I just took 7 and 5 as a factor nothing else and then I multiply 1 ex. and 2ex. and found that in the last expression taking 7 as common looks the same then those got cancel and end up with 10/7.
I'm not sure how you would prove it, but
Given x+y+z=0 and (Sum from i=0 to i=n (a_i)) =(Sum from j=0 to j=m (b_j))
where each a_i and b_i is a prime>1.
Then
(Product from i=0 to i=n (x^(a_i)+y^(a_i)+z^(a_i))) /(Product from j=0 to j=m(x^(b_j)+y^(b_j)+z^(b_j))) =(Product from i=0 to i=n (-1^ceiling(a_i/2)*a_i)) /(Product from j=0 to j=m(-1^ceiling(b_j/2)*b_j))
Wow!
@@SyberMath It does make it kind of easy to setup these kinds of questions.
11:37 you can just factor it as a square of a trinomial
Yes!
This method is so lengthy ,by the concept of theory of equations one can do with in 5 minutes by forming simple related equalities.
Dude! This problem was so complex, if I had tried this before the video I wud hav used up like 10 pages and realize none of the approach works, and it becomes so complex that my noob brain can't handle it
Put x=2, y=-1 z=-1 and the expression value comes out 10/7. It's a 15 secs sum which generally come in aptitude test.
Observe that the given problem is equivalent to sum of roots of cubic eqation in t as 0 .Now use vietas method and recurrence to get it sorted out fast ! Anyway ur method was also nice
Your method is very nice, too!
@@SyberMath Thanks brother ❤️
Your method is slick bro!!! If S[n]=x^n+y^n+z^n then the problem becomes S[2]*S[5]/S[7].... using vieta's and forming a recursion involving S[n] I had letters cancel out and I only bothered about S[2] which is -2(xy+yz+xz) or for me is -2b since b is the coefficient of x in the cubic equation....involving recursion, the rest was history
use Newton's identities for power sums and elementary symmetric polynomials.
for 3 variables (the roots of some cubic):
e1=p1=x+y+z=0
e2=xy+xz+yz
e3=xyz
p2=(e1)(p1)-2(e2) = -2(e2)
for j>2: pj = (e1,-e2,e3) dot (pj-1,pj-2,pj-3) and using p1=e1=0:
p3=3(e3)
p4=2(e2)^2
p5=-5(e2)(e3)
p6=-2(e2)^3 + 3(e3)^3
p7=7(e2)^2(e3)
therefore:
(p2)(p5)/(p7) = (-2(e2))(-5(e2)(e3))/[7(e2)^2(e3)]
10(e2)^2(e3)/[7(e2)^2(e3)]
= 10/7
[]
Interesting... In the problem the exponents are [2,5,7]. The only other solutions I can find with similar behaviour are [2,3,5] and [3,4,7].
the first 2 #s add to the last
Assume x=2, y=1, z=1; answer is 10/7
(-2,1,1)
But first of all you must prove that it is a constant.
I observed one thing 2+5=7
nice fact lmao
@@MathElite 😆
@@akshatjangra4167 hello how are you?
@@imagoat9070 i am good, and as i can see we three are together again
@@akshatjangra4167 ❤️
Incredible solution bro.
Thank you! 💖
Please never skip a nontrivial step.
You're right
Now that was really tough to solve. I didn't try this out, because I thought I have to raise x+y+z to the 5th and 7th powers... That probably wouldn't help me at all, solving this problem required quite a lot of invention. Well, you definitely made up for this "one liner" from a few days ago, where result was -3 and the video could have lasted 1 minute. : ) Keep up the good content.
Thank you! The point for that video was how painful a solution can be if you do not follow some shortcuts or some elegant method. I like introducing two methods because everyone learns differently, it gives a chance to people to compare and contrast, and it's good for improving problem solving skills.
@@SyberMath Lovely. I appreciate it. I also appreciate that despite your channel is growing, you still read every comment under the videos. That's important.
@@snejpu2508 💖
Really great video! I like this simplification problems!
Great to hear!
Ah! Beauty...Expressed in symbols💙
💖
Thank you for your efforts ... You have good videos and good presentation of problem solutions. I hope that your channel will continue to grow. Develop further and you will succeed. I wish you all the best ... See you soon ...
So nice of you 💖
Intersting ! Good job !
Thanks!
Well interesting and fun
very interesting solution sir thank u
Most welcome
A surpring result!
I’m these problems typically if you want to solve fast you can assume that plugging any number won’t make any difference ( unless you plug in a value that makes the denominator undefined )
Plug in -1 and -1 and 2 and you will get :
1+1+4=6
-1+-1+32=30
180/(-1+-1+128) =180/126=20/14=10/7
Done!
*in these problems
This is super nice.
I appreciate that! 😊
Nice and quick challenge=> what is/are the values of sqrt4
2 only 2 not negative 2 but 2
@@MathElite ABSOLUTEly correct,and pun intended
2
@@advaykumar9726 yes
Why is it just 2 and not -2??
it can be solved by assuming x=-1 y=-1 and z=2 for example
Maybe start with plotting to find its constant and then...
Should just 1 point (or is it line or surface) omit, where x^7+y^7+z^7=0 ? (To avoid division by zero)
The validity at division by zero could be possible proven with the limes testing.
If for some reason the original equation is not important here this should be correct.
I have not done it but got the answer within a minute. From the question it is clear that the answer will be a number. Then put x = 1/2 =y, z = -1.
so we have x + y + z = 0 and the powers of 2, 5 and 7 and get 2*5/7. Does that mean that the powers of a, b and c (all naturals) will result in a*b/c?
Good question! I doubt it but let's hear from other people
If (x^a+y^a+z^a)(x^b+y^b+z^b)/(x^c+y^c+z^c) can be simplified to a constant under the condition x+y+z=0, then the value must equal what you get if you take (x, y, z) = (1, -½, -½), which is one point satisfying the condition. The value then becomes (1+2(-½)^a)(1+2(-½)^b)/(1+2(-½)^c), which does not equal ab/c in the general case.
@@md2perpe you're probably right. It's easy to check it out but I'm not tempted cuz it doesn't look like there's an elegant solution for this problem (unless induction will do the trick): it was solved by brute force
Let xⁿ+yⁿ+zⁿ=P(n)
(x+y+z)²=0, x²+y²+z²=-2(xy+xz+yz)
(x+y+z)(x²+y²+z²-xy-xz-yz)=x³+y³+z³-3xyz=0, x³+y³+z³=3xyz
Using the identity P(n)=P(n-1)(x+y+z)-P(n-2)(xy+xz+yz)+P(n-3)xyz=P(n-2)P(2)/2+P(n-3)P(3)/3
P(4)=P(2)²/2
P(5)=P(2)P(3)/2+P(2)P(3)/3=5/6(P(2)P(3))
P(7)=5/12(P(2)²P(3))+1/6(P(2)²P(3))=7/12(P(2)²P(3))
P(2)P(5)/P(7)=(5/6P(2)²P(3))/(7/12P(2)²P(3))=10/7
ab/c hypotesis doesn't work because if we substitute P(3)P(4)/P(7) yielding 6/7 instead of 6/7
Wow!
Just amazing . . .
Thank you!
amazing
Incredible I love it. I'm alwàys wonderinf if There is concrete application in physics for instance
Thank you! 💖
@@SyberMath I guess that x+y+z=0 or x^n+y^n+z^n = 0 and [(x^{n+1}+y^{n+1}+z^{n+1})(x^{n+3}+y^{n+3}+z^{n+3})]/(x^{2n+4}+y^{2n+4}+z^{2n+4})=C I'll check with google sheet
a) The solution is 10/7, if x ≠ 0 and y ≠ 0 and z ≠ 0.
b) There is no solution, if x = 0 or y = 0 or z = 0.
0/0 is indeterminate form therefore no solutions
Good problem.
Yes, thanks
x+y+z = 0 implies
x^2 + y^2 + z^2 = -2(xy+yz+ zx)
Nice tx
No problem
x+y+z = 0 ; if x or y or z = 0, we cannot find the answer.
x+y+z=0, (x+y+z)^7=0, the expression=0
Symetrial polynomials and basic theorem so the nominator and denominator can be written in polynomial of (x+y+z)^n + something. Due condition all is about what last number of linear term is in nominator and denominator and this is answer. When we see it better the answer is possible to see inmediatelly :) Too easy.
Which basic theorem are you referring to?
So how you can generalize this statement? For which n and m (x^n+y^n+z^n)*(x^m+y^m+z^m)/(x^(n+m)+y^(n+m)+z^(n+m)) is independent of the choice of x, y, z?
Put x=y=-1 and z,=2 simple
@@sakshamsingh1778 And how does this prove the theorem in general?
@@sakshamsingh1778 lmao
So you get 10/7, provided x -y and x^2+xy+y^2 0.
or substitute x=1,y=-1,z=0 and x=2,y=-1,z=-1 and see that the answer is always the same
You get 0/0 in the first case.
@@khundeejai7945 true but I think you get my point ;-)
You are so awesome 😎😎😎😎😎😎😎😎😎 and a pro!
You too
@@SyberMath thanks for your videos! Take care and lots of support!
x=1 , y=1, z=-2 (then x+y+z=0) 대입... end!
😁
I was watching art videos and then youtubes algorithm knew it had to show me this
🙂
Using the results of another video you only need to find out what (x²+y²+z²)(x³+y³+z³) is, however I get -10/7. Either I made a mistake or you.
Ok, found it. ;-)
😁
I have a question: Are there infinitely many solutions for this equation? (15)^2a+15b+c=13
If there is a finite number of solutions: which are they?
And if there are infinitely many solutions and I use other equations would there be less solutions?
Please help me I need this because I want to make a graph in an unusual way...
Thanks
There are infinite real solutions. Leave a free,
Set b=-15^(2a-1), set c=13.
Leave a free, set b=-15^(2a-2) set c=13-14*15^(2a-1)
etc.
@@ThAlEdison Thanks a lot!
x=1 y=1 z=-2
x+y+z=0
...... =10/7
2 (X^5+ y^5+z^5) + 2( x^2+ y^2+z^2) what is this =0 Since
or X+y + z =0/27 = 0 gojamil onko
Let x=y=1 and z=-2 we get x+y+z=0 and 10/7 is the value of the expression!
You cant just substituing some value for x,y,z because we need to prove for all x,y,z satisfy x+y+z=0
But how could I know, without proofs, that it's really a constant?