a-1/a = b : a/b-1/ab =1 .......1 b-1/b = c : b/c-1/bc =1.........2 c-1/c = a : c/a-1/ca = 1.......3 adding 1, 2 and 3 and rearranging we get a/b+b/c+c/a = 3 + 1/ab +1/bc + 1/ca .......4 1/a = a-b : 1/b= b-c : 1/c=c-a from given sunstituting all these in 4 we get a(b-c)+b(c-a)+c(a-b) = 3+1/ab+1/bc+1/ca expanding we get ab-ca+bc-ab+ca-bc = 3+1/ab+1/bc+1/ca 0 = 3 + 1/ab+1/bc+1/ca rearranging we get 1/ab+1/bc+1/ca = -3
a-1/a=b can be written as a^2-1=ab => 1/ab=1/a^2-1 similarly if we do for other 2 equations fallowed by adding them and rearranging can get the same solution without using algebraic formulae.
1/a = a-b; 1/b=b-c; 1/c=c-a; ab=a^2-1; bc=b^2-1; ca=c^2-1. 1/(ab)+1(bc)+1(ac)= (a-b) (b-c) +(b-c) (c-a) +(a-b) (c-a) = ab+bc+ca-(a^2+b^2+c^2)=-3. Thank you for your work, Sir. I amuse with your questions.
Would be helpful to talk about the symmetry that motivates your algebra choices. Otherwise it looks like a magic trick that students cannot apply to other problems. Very good channel content overall!
Here is my way (without looking the video): summing also equations we get 1/a+!/b+1/c=0! if we do square on the initial set of equations : a^2+1/a^2-2=b^2; b^2+1/b^2=c^2-2; c^2+1/c^2-2=a^2; Summing the equations we get 1/a^2+1/b^2+1/c^2=6. since 1/a+1/b+1/c=0; => [1/a+1/b+1/c]^2=0 too. or 1/a^2+1/b^2+1/c^2+2(1/ab+1/bc'1/ac)=0 at the end we get 1/ab+1/bc'1/ac=-3
🤔hmmmm....a perfect-square trinomial...(a+b+c)^2...a perfectly square tri-anything, is an oxymoron. In this case though the oxymoron is a paradox because (a+b+c)^2 is absolutely congruent...🙂
For a problem like this, just going through the mechanics of the solution isn't going to be conducive to learning how to solve problems like this. There needs to be explanation as to what inspired each of these steps and the reasoning behind them. Otherwise the solution seems to fall out of thin air.
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a-1/a = b : a/b-1/ab =1 .......1
b-1/b = c : b/c-1/bc =1.........2
c-1/c = a : c/a-1/ca = 1.......3
adding 1, 2 and 3 and rearranging we get
a/b+b/c+c/a = 3 + 1/ab +1/bc + 1/ca .......4
1/a = a-b : 1/b= b-c : 1/c=c-a from given
sunstituting all these in 4 we get
a(b-c)+b(c-a)+c(a-b) = 3+1/ab+1/bc+1/ca
expanding we get
ab-ca+bc-ab+ca-bc = 3+1/ab+1/bc+1/ca
0 = 3 + 1/ab+1/bc+1/ca
rearranging we get
1/ab+1/bc+1/ca = -3
Excellent!
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Excellent approach! Thank you for sharing your method!
Excellent question! Brilliant Teacher!!!!!!👍
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this multivariable rational equation took a lot of steps, very well done, thanks for sharing
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Can you make more such Olympiad videos? Because I am preparing for Olympiads and your questions and their explanation is amazing.
Sure! More to come...
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Frumos! Cu artificii cu tot tacamul! Super! Next?!
Greetings from Thailand i always found you video on recommended and the problem look interesting and i like it!😊
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Such amazing efforts by you sir! Keep it up! Our math family will grow soon!
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a-1/a=b can be written as a^2-1=ab => 1/ab=1/a^2-1 similarly if we do for other 2 equations fallowed by adding them and rearranging can get the same solution without using algebraic formulae.
1/a = a-b; 1/b=b-c; 1/c=c-a; ab=a^2-1; bc=b^2-1; ca=c^2-1.
1/(ab)+1(bc)+1(ac)= (a-b) (b-c) +(b-c) (c-a) +(a-b) (c-a) = ab+bc+ca-(a^2+b^2+c^2)=-3.
Thank you for your work, Sir. I amuse with your questions.
How did you know squaring will help in solving the problem ?
May be trail n error... The only time tested successful method
You are helping us from developing Alzheimer’s 😊
Classic working. It will not occur to all the way it occurred to Professor. Many will take a long time to solve.
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Thank u so much profecor please what is the program u are using to explain the problem??
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Would be helpful to talk about the symmetry that motivates your algebra choices. Otherwise it looks like a magic trick that students cannot apply to other problems. Very good channel content overall!
Great suggestion!
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BEAUTIFUL QUESTION
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Very good solution ,dear professor
math has been always the toughest for me
you have made it easy
Excellent effort
Here is my way (without looking the video):
summing also equations we get 1/a+!/b+1/c=0!
if we do square on the initial set of equations :
a^2+1/a^2-2=b^2; b^2+1/b^2=c^2-2; c^2+1/c^2-2=a^2;
Summing the equations we get 1/a^2+1/b^2+1/c^2=6.
since 1/a+1/b+1/c=0; => [1/a+1/b+1/c]^2=0 too.
or 1/a^2+1/b^2+1/c^2+2(1/ab+1/bc'1/ac)=0
at the end we get 1/ab+1/bc'1/ac=-3
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Namaste sirji.
Thanks for video. Good luck!!!!!!!
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Sir very helpful content
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Best solution thanks a lot 🌹🌹🇮🇳👍
Very well explained
WOW!! This was a cool problem! Thanks!
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great problem
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Where did u used eq 1?
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Very simple and easy calculation 🧮🧮🧮🧮🧮 great 👍👍👍👍👍 sir ❤️🙏🙏🙏
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It is a nice solution, but I don't feel I learned anything from it. What is the hint that squaring the equations will help?
🤔hmmmm....a perfect-square trinomial...(a+b+c)^2...a perfectly square tri-anything, is an oxymoron. In this case though the oxymoron is a paradox because (a+b+c)^2 is absolutely congruent...🙂
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Nice explanation 🙂
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Really helpful thnx a lot
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thank you
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Super!!!
So any problem have solution
nice solution
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Idk why it is smooth but it is smooth. Good question.
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For a problem like this, just going through the mechanics of the solution isn't going to be conducive to learning how to solve problems like this. There needs to be explanation as to what inspired each of these steps and the reasoning behind them. Otherwise the solution seems to fall out of thin air.
I didn’t solve the problem but by seeing your results I concluded abc=-1
Solved easily
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Continuez comme ça 👍
Hhh vous êtes français ? Je regarde ce vidéo car je suis entrain de préparer pour les olympiades des mathématiques qu'ils seront aujourd'hui
Marocaine ici!
@@uchihataha8643 Très peu de français.
Tout le meilleur cher.
V nice
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Magical
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-3
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Again nostralgic.
I feel ,again I am in class nine.
Which is 31 years back
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3 i must have messed up the sign
Three thumbs up, each thumb > 0 :-).
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