Very nice video and long calculus too. It is a surprise to have such nice simple solution without (x+y) in fact. If you name Q(x,y) the original expression, you can verify that Q(nx,ny)=n².Q(x,y). So the final expression must lead the same finding which IS n'ont the case with Factor (x+y).
Interestingly, with the numerator using powers of 5 and the denominator 3, the solution is 5/3 * (x² + xy + y²). This is the only other occasion where it is of this form.
I got a different answer: (7/5) (x^2 + xy + y^2). When I use a computer program and use various non-zero values for x and y, my answer checks with the original equation. Yours with the extra (x+y) does not. **EDIT: lots of people saw the same thing. I agree with the comments section, there's an (x+y) in the denominator, so that cancels out of the solution.
I really like SyberMath but I think there's an issue here. I used MATLAB and get something different. The first thing that bugged me was dimensions. If x and y have units of meters, the units of the original fraction must have units of m^2. The solution at the end video would have units of m^3. I get (7/5)(x^2 + xy + y^2) The next thing to check is that if a result is true in general it's true for any specific case. So let's test, x = 1, y=1, that would make the original function (2^7-2)/(2^5 - 2) = 126/30 = 4.2 SyberMath's answer 1.4*2*3 = 8.4 my answer 1.4*3 = 4.2 What do you think?
@@faiselbutt2944 Yep! I checked it out using MathCAD, same thing. Using (7/5) (x^2 + xy + y^2) and assigning various real values for x and y, it checks out numerically with the original equation. With the extra (x+y) in the answer, the numbers don't work out. (Note: x and y cannot be zero or you get a divide-by-zero problem.)
The Lesson: the risk of leaving a part out is that you might forget to put it back in at the end. With a large piece of physical paper you can put a visible reminder somewhere, but with electronic "paper" that scrolls away, it is difficult to leave oneself such a reminder.
We will count. that the fraction is reduced, giving, as a result, a polynomial without remainder in the form of another fraction. Then. [(x+y)^7 -x^7-y^7]/[(x+y)^5 -x^5 -y^5] = Ax^2 +Bxy+Cy^2, where constants A,B and C must be determined. This equality must hold for any x and y. We assume in it sequentially, for example, y=x, y=x/2 and y=2x. After reducing such terms, we obtain a system of linear equations with respect to A,B and C: A+B+C=21/5, A+B/2+C/4=49/5 , A +2B+4C=49/20, solving which, we get A=B=C=7/5. Answer: 7/5 (x^2+xy+y^2).
The equation y⁷ = (x + y)⁷ - x⁷ is equivalent to x⁷ + y⁷ = (x + y)⁷. It's not difficult to proof that (x + y)⁷ = x⁷ + y⁷ + 7xy(x + y)(x² + xy + y²)². So, x⁷ + y⁷ = (x + y)⁷ is the same that xy(x + y)(x² + xy + y²)² = 0. From this result you get three equations: (x = 0) ∨ (y = 0) ∨ (x + y = 0). This is why the graph of y⁷ = (x + y)⁷ - x⁷ only has straight lines.
This would have helped simplify: xⁿ- yⁿ = (x - y)(xⁿ⁻¹ + xⁿ⁻²y + … + xyⁿ⁻² + yⁿ⁻¹) If n is odd (as is the case in the numerator and denominator) and substitute y with -y: xⁿ + yⁿ = (x + y)(xⁿ⁻¹ - xⁿ⁻²y + … - xyⁿ⁻² + yⁿ⁻¹) But yes, you needed to cancel out the x+y in the video.
Very curious about a creepy thing here . . . A 7th degree polynomial divided by a 5th degree polynomial ends up with a 3rd degree polynomial (!!) How is that even possible?
I did forget to cancel out the x+y term! Oooopsies! 😮😮
Weird that a division of a 7 degree polynomial with a 5 degree one gives a 3 degree polynomial.😊
Fault confessed is half redressed !
Faute avouée est à moitié pardonnée !
You forgot an (x+y) in the denominator, which simplifies with the one in the numerator
Uh-oh! 😮
the right answer should be. 7/5 * (x^2 + xy + y^2)
Umm you forgot to put the x+y in the final denominator which cancel out the x+y from the numerator
Uh-oh! 😮
Very nice video and long calculus too. It is a surprise to have such nice simple solution without (x+y) in fact. If you name Q(x,y) the original expression, you can verify that Q(nx,ny)=n².Q(x,y). So the final expression must lead the same finding which IS n'ont the case with Factor (x+y).
Thanks for sharing
Interestingly, with the numerator using powers of 5 and the denominator 3, the solution is 5/3 * (x² + xy + y²). This is the only other occasion where it is of this form.
I got a different answer: (7/5) (x^2 + xy + y^2). When I use a computer program and use various non-zero values for x and y, my answer checks with the original equation. Yours with the extra (x+y) does not.
**EDIT: lots of people saw the same thing. I agree with the comments section, there's an (x+y) in the denominator, so that cancels out of the solution.
I really like SyberMath but I think there's an issue here. I used MATLAB and get something different. The first thing that bugged me was dimensions. If x and y have units of meters, the units of the original fraction must have units of m^2. The solution at the end video would have units of m^3. I get
(7/5)(x^2 + xy + y^2)
The next thing to check is that if a result is true in general it's true for any specific case. So let's test, x = 1, y=1, that would make the
original function (2^7-2)/(2^5 - 2) = 126/30 = 4.2
SyberMath's answer
1.4*2*3 = 8.4
my answer
1.4*3 = 4.2
What do you think?
He forgot an (x+y) in the denominator.
@@faiselbutt2944
Yep! I checked it out using MathCAD, same thing. Using (7/5) (x^2 + xy + y^2) and assigning various real values for x and y, it checks out numerically with the original equation. With the extra (x+y) in the answer, the numbers don't work out. (Note: x and y cannot be zero or you get a divide-by-zero problem.)
The denominator factorization is missing a (x+y)
The result is degree 3 and denominator is degree 5 ---> degree (result * denominator) = 8 ---> error somewhere !
You forget to cancel out x+y in the numerator. I got (7/5)*(x^2+xy+y^2)
When I did it, I got (7/5)*(x2+xy+y2)...
I think you forgot a factor (x+y) in the denominator,which cancels the same factor in the numerator...
Great. But (x+y) was forgotten in denominator. So the (x+y) factor disapper in the given solution. This made me think of Lamé.
Some careless mistakes in omitting the common factor (x+y) in denominator. By cancellation , the final result is 7/5 (x^2+xy+y^2)#
i try this question and i got my answer
but where is (x+y) on my solution??????
omg
Sir how could we request for solving question
The Lesson: the risk of leaving a part out is that you might forget to put it back in at the end. With a large piece of physical paper you can put a visible reminder somewhere, but with electronic "paper" that scrolls away, it is difficult to leave oneself such a reminder.
We will count. that the fraction is reduced, giving, as a result, a polynomial without remainder in the form of another fraction. Then.
[(x+y)^7 -x^7-y^7]/[(x+y)^5 -x^5 -y^5] = Ax^2 +Bxy+Cy^2, where constants A,B and C must
be determined. This equality must hold for any x and y.
We assume in it sequentially, for example, y=x, y=x/2 and y=2x.
After reducing such terms, we obtain a system of linear equations with respect to A,B and C:
A+B+C=21/5, A+B/2+C/4=49/5 , A +2B+4C=49/20,
solving which, we get A=B=C=7/5.
Answer: 7/5 (x^2+xy+y^2).
Even Honer sometimes nods.
you forgot the factor of x + y in the denomator
For some reason i plotted the graph of y^7=(x+y)^7-x^7. I got the graph to be linear. Can you tell why?
The equation y⁷ = (x + y)⁷ - x⁷ is equivalent to x⁷ + y⁷ = (x + y)⁷. It's not difficult to proof that (x + y)⁷ = x⁷ + y⁷ + 7xy(x + y)(x² + xy + y²)². So, x⁷ + y⁷ = (x + y)⁷ is the same that xy(x + y)(x² + xy + y²)² = 0. From this result you get three equations: (x = 0) ∨ (y = 0) ∨ (x + y = 0). This is why the graph of y⁷ = (x + y)⁷ - x⁷ only has straight lines.
u missed the x+y term in denominator just checkout.
mmm heptic divide by quintic should be quadratic...
You could also represent the solution as 7/3 * (x^3 + 2*x^2*y + 2*x*y^2 + y^3)
You have forgotten (x+y) in denominator... so answer should be 7/5(x^2 +xy +y*2) only!!
👍
myLifeTime = myLifeTime - 5
power 7 - power 5 = power 2 , we have in the solution power 3 . something is wrong
because I forgot to write x+y in the denominator which cancels out
This would have helped simplify:
xⁿ- yⁿ = (x - y)(xⁿ⁻¹ + xⁿ⁻²y + … + xyⁿ⁻² + yⁿ⁻¹)
If n is odd (as is the case in the numerator and denominator) and substitute y with -y:
xⁿ + yⁿ = (x + y)(xⁿ⁻¹ - xⁿ⁻²y + … - xyⁿ⁻² + yⁿ⁻¹)
But yes, you needed to cancel out the x+y in the video.
Where is the 2nd method??😂😂
😁
You get 0 marks for your answer.
Sorry teacher! 😄
Very curious about a creepy thing here . . . A 7th degree polynomial divided by a 5th degree polynomial ends up with a 3rd degree polynomial (!!)
How is that even possible?
I think something wrong here🤔