I solved it slightly differently. SInce a, b and c are distinct, I set a = b+h and b=c+k, with h and k rational numbers. So (a-b)^2 = h^2, (b-c)^2=k^2 and (c-a)^2 = (h+k)^2. The original expression becomes sqrt(1/h^2+1/k^2+1/(h+k)^2). Cross multiplication and simplification brings to sqrt((k^4+h^4+3(hk)^2+2hk(k^2+h^2))/(hk(h+k))^2. The denominator is a perfect square. The numerator can be rearranged to (k^2+h^2+hk)^2, which is alo a perfect square. Since numerator and denominator are sum and products of the rationals h and k, they are also rational. SInce numerator and denominator are rational, the fraction is rational, which is what we had to demonstrate
here’s another method : observe [1/(a-b) + 1/(b-c) + 1/(c-a)]^2. this, upon expanding, turns into 1/(a-b)^2 + 1/(b-c)^2 + 1/(c-a)^2 + 2[1/(a-b)(b-c) + 1/(b-c)(c-a) + 1/(c-a)(a-b)] focus on that last part for a bit. upon multiplying each term by (c-a), (a-b) and (b-c) respectively, we can see that the numerator of that part will become (c-a+a-b+b-c) or zero, for short. hence that entire term disappears. as a result, we have shown that [1/(a-b) + 1/(b-c) + 1/(c-a)]^2 = 1/(a-b)^2 + 1/(b-c)^2 + 1/(c-a)^2. take the square root on both sides, and you’ll see that on the LHS, due to it being a square, the square and square root will cancel out, hence making it rational, given a, b and c themselves are rational.
I solved it a little differently... I got to the part where you needed to prove the numerator (MN)^2 + (M+N)^2N^2 + (M+N)^2M^2 was a perfect square, but from there I multiplied out everything directly. I'll spare you the messy details but basically it amounts to M^4 + 2M^3N + 3M^2N^2 + 2MN^3 + N^4. And then another trick: divide everything by N^4. N cannot be 0 and N^4 is a perfect square, so this is a perfectly valid thing to do. I end up with (M/N)^4 + 2(M/N)^3 + 3(M/N)^2 + 2(M/N) + 1. One last substitution: F = M/N, and now I have a polynomial! F^4 + 2F^3 + 3F^2 + 2F + 1 Now I can easily factor this by inspection: (F^2 + F + 1)^2. Perfect square achieved.
Previously the thumbnail used to stay for 1 or 2 seconds at the starting of video but now it only stays for an instance. Please let the problem stay for at least a second at the beginning , sir.
Reasonable suggestion! 👍 I'm watching from a PC in a Chrome browser and I've recently came across a TH-cam Autostop extension - now a new video no longer autostarts and I can take a look at the problem and try to solve it before I watch the video.
Question request for explanation: Let f be a polynomial function that satisfies f(x)+f(x/y^2)+f(x/y)=f(x)*f(1/y)-(1/y^3)+(x^3/y^6)+2 for all x belongs to R-{0},f(1)≠1,f(2)=9. Find the value of r=1->100Σ(f(r))
i have an idea to prove m^2n^2+l^2n^2+l^2m^2 is a perfect square. we can add and subtract (2mln^2+2m^2ln+2ml^2n) so after adding this trem we get a perfect square that is (mn+ln+lm)^2. (x+y+z)^2 is (x^2+y^2+z^2+2xy+2yz+2zx) so after this subsitution we get our equation as (mn+ln+lm)^2 -(2mln^2+2ml^2n+2m^2ln). after this we can take -2mln common on the other term so we get it as (mn+ln+lm)^2- 2mln(l+n+m). Since l+m+n is zero so we get (mn+ln+lm)^2 which is a perfect square
It has to apply to every instance, not just one case. For example 4 + 16 = 20 When you try to prove your theories don't argue with an example for it, instead try to find an example against it - if you can't, you are probably right. It's how science has been rolling for a looong time
I proved it differently. I added the fractions. The number with the sum of three squares which can be written as [(b-c)(c-a)+(a-b)(c-a)+(a-b)(b-c)]^2 - 2(a-b)(b-c)(c-a)[0]. Since the numerator and denominator are perfect squares and a,b,c are rational numbers, the given expression is rational. I solved this on paper in around 3 minutes using this method.
Sir, you have got an extraordinary teaching skills coz you are the authority of this subject.Hats off to you.
Not only algebra skill teaching skill is also amazing. Thanks to u.
I solved it slightly differently. SInce a, b and c are distinct, I set a = b+h and b=c+k, with h and k rational numbers. So (a-b)^2 = h^2, (b-c)^2=k^2 and (c-a)^2 = (h+k)^2.
The original expression becomes sqrt(1/h^2+1/k^2+1/(h+k)^2). Cross multiplication and simplification brings to sqrt((k^4+h^4+3(hk)^2+2hk(k^2+h^2))/(hk(h+k))^2. The denominator is a perfect square. The numerator can be rearranged to (k^2+h^2+hk)^2, which is alo a perfect square. Since numerator and denominator are sum and products of the rationals h and k, they are also rational. SInce numerator and denominator are rational, the fraction is rational, which is what we had to demonstrate
here’s another method :
observe [1/(a-b) + 1/(b-c) + 1/(c-a)]^2.
this, upon expanding, turns into 1/(a-b)^2 + 1/(b-c)^2 + 1/(c-a)^2 + 2[1/(a-b)(b-c) + 1/(b-c)(c-a) + 1/(c-a)(a-b)]
focus on that last part for a bit. upon multiplying each term by (c-a), (a-b) and (b-c) respectively, we can see that the numerator of that part will become (c-a+a-b+b-c) or zero, for short. hence that entire term disappears.
as a result, we have shown that [1/(a-b) + 1/(b-c) + 1/(c-a)]^2 = 1/(a-b)^2 + 1/(b-c)^2 + 1/(c-a)^2. take the square root on both sides, and you’ll see that on the LHS, due to it being a square, the square and square root will cancel out, hence making it rational, given a, b and c themselves are rational.
I'm always happy to see your videos so I don't hesitate to tap and watch 😊❤🎉. You're such a an amazing teacher
I solved it a little differently... I got to the part where you needed to prove the numerator (MN)^2 + (M+N)^2N^2 + (M+N)^2M^2 was a perfect square, but from there I multiplied out everything directly.
I'll spare you the messy details but basically it amounts to M^4 + 2M^3N + 3M^2N^2 + 2MN^3 + N^4.
And then another trick: divide everything by N^4. N cannot be 0 and N^4 is a perfect square, so this is a perfectly valid thing to do.
I end up with (M/N)^4 + 2(M/N)^3 + 3(M/N)^2 + 2(M/N) + 1.
One last substitution: F = M/N, and now I have a polynomial!
F^4 + 2F^3 + 3F^2 + 2F + 1
Now I can easily factor this by inspection: (F^2 + F + 1)^2.
Perfect square achieved.
I was going to use a brute force attack, applying the differences as written. Your way of writing parametric equations first is much more elegant.
Thank you,again. This is the many beauties of mathematics ❤
Previously the thumbnail used to stay for 1 or 2 seconds at the starting of video but now it only stays for an instance. Please let the problem stay for at least a second at the beginning , sir.
Reasonable suggestion! 👍
I'm watching from a PC in a Chrome browser and I've recently came across a TH-cam Autostop extension - now a new video no longer autostarts and I can take a look at the problem and try to solve it before I watch the video.
you can turn it off in the settings
Excellent question😄😄
Very clever algebraic manipulations!
Question request for explanation:
Let f be a polynomial function that satisfies f(x)+f(x/y^2)+f(x/y)=f(x)*f(1/y)-(1/y^3)+(x^3/y^6)+2 for all x belongs to R-{0},f(1)≠1,f(2)=9.
Find the value of r=1->100Σ(f(r))
Practice makes the man perfect.
Wonderful 🌸🌺🌻🌹🌷🌼💐
The numerator of the radicand becomes (lm)^2 + (ln)^2 + (mn)^2. consider q = (lm) + (ln) + (mn). Hence q^2 = (lm)^2 + (ln)^2 + (mn)^2 + 2(lmn)(l + m + n). Note that (l + m + n) = 0. Thus the numerator is q^2, a perfect square. QED
Great tricks
That was beautiful to watch, tysm.
Very good. Thanks 👍
❤❤❤❤❤❤❤ so good to learn this algebra substitution.. can i know where can i buy the t shirt with infinity logo on it...
nice one!
Genius ❗️
i have an idea to prove m^2n^2+l^2n^2+l^2m^2 is a perfect square. we can add and subtract (2mln^2+2m^2ln+2ml^2n) so after adding this trem we get a perfect square that is (mn+ln+lm)^2. (x+y+z)^2 is (x^2+y^2+z^2+2xy+2yz+2zx) so after this subsitution we get our equation as (mn+ln+lm)^2 -(2mln^2+2ml^2n+2m^2ln). after this we can take -2mln common on the other term so we get it as (mn+ln+lm)^2- 2mln(l+n+m). Since l+m+n is zero so we get (mn+ln+lm)^2 which is a perfect square
Sir I request you to search for JEE Question papers which will provide you numerous number of amazing problems.
Very Diophantine :)
So savage ❤️❤️
You said the sum of squares is not a square but 16 + 9 = 25 which is a square?
You know what I meant
It has to apply to every instance, not just one case. For example
4 + 16 = 20
When you try to prove your theories don't argue with an example for it, instead try to find an example against it - if you can't, you are probably right. It's how science has been rolling for a looong time
Hm, but are there two cubes that sum up to a cube? 🤔
😂😂😂 Fermat’s last theorem …
Am I the only one that hears the outro: never stop learning, never stop learning, stop living. 😆
It is, never stop learning, those who stop learning stop living
I proved it differently. I added the fractions. The number with the sum of three squares which can be written as [(b-c)(c-a)+(a-b)(c-a)+(a-b)(b-c)]^2 - 2(a-b)(b-c)(c-a)[0]. Since the numerator and denominator are perfect squares and a,b,c are rational numbers, the given expression is rational. I solved this on paper in around 3 minutes using this method.
First
As we say in portuguese, osso 🦴, It meens a Very difficult situation!