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 RU-vid 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.
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.
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.
