Note that this technique is particularly effective when, as here, x times it's algebraic conjugate equals 1; so x is a unit in the relevant adjunct field Z[...].
Brute force works too (you get (9900-5720sqrt3)/(26-15sqrt3) and later -220sqrt3) But as usual the elegant PreMath solution is way better! Thanks as always for the fun daily puzzle!
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Thanks!!! That was a amzing way to do it, and i love it! Keep up your good work dear my freind Sir PreMath, and i wish you all the best. love and prayers from YoonHo!
So nice of you, YoonHo Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀
I went about this in a sligthly different way. From the first equation, I calculated X²=7-4sqr(3), X²-X+1=3X and X²+X+1=5X. From the numerator of the second I got 7X⁷+5X⁵-5X³-7X =7X(X³+1)(X³-1)+5X³(X²-1)=7X(X+1)(X-1)(X²+X+1)(X²-X+1)+5X²(X²-1). Semplificatin with the denominator (since X is not 0) leads to [7(X²-1)(X²+X+1)(X²-X+1)+5X²(X²-1)]/X³. From here using the first equation I get (X²-1)(7*3X*5X+5X²)/X³=110X²(X²-1)/X³=110(X²-1)/X. Subsitituing here the values of X and X², and rationalizing the result, brings to the result.
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Fantastic as usual!
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Note that this technique is particularly effective when, as here, x times it's algebraic conjugate equals 1; so x is a unit in the relevant adjunct field Z[...].
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@@PreMath Thank YOU!
The most tricky part is to use the fact that the product of x and its conjugate equals 1. Nice question and wonderful step-by-step explanation!👍
Thank you interesting exercice
Elegant and efficient
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Brute force works too (you get (9900-5720sqrt3)/(26-15sqrt3) and later -220sqrt3) But as usual the elegant PreMath solution is way better! Thanks as always for the fun daily puzzle!
Thank u sir, pls don't stop teaching these important problems
Hope u have a great one 😊
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Thanks!!! That was a amzing way to do it, and i love it! Keep up your good work dear my freind Sir PreMath, and i wish you all the best. love and prayers from YoonHo!
So nice of you, YoonHo
Glad to hear that!
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@@PreMath Thanks!!!
Great explanation👍
Thank you so much for sharing😊😊
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this is excellent prep for a Math Olympiad, excellent job
Great!
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Nice problem. I factored first and divided later. Not sure if that is easier or harder.
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Thanks for video.Good luck sir!!!!!!!!!!!
Did it by factorising differently and getting values for X, 1/x,x^2 and 1/ x^2 ending with -220 times root 3.
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It's a awsome question with a awsome answer and very nice explanation👍. Thank you teacher 🙏.
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Fantastic. Excellent working.
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Great video! I'm going to try these
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V nice way of explanation
Thanks for liking, Niru
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Lovely problem
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I went about this in a sligthly different way. From the first equation, I calculated X²=7-4sqr(3), X²-X+1=3X and X²+X+1=5X. From the numerator of the second I got 7X⁷+5X⁵-5X³-7X =7X(X³+1)(X³-1)+5X³(X²-1)=7X(X+1)(X-1)(X²+X+1)(X²-X+1)+5X²(X²-1). Semplificatin with the denominator (since X is not 0) leads to [7(X²-1)(X²+X+1)(X²-X+1)+5X²(X²-1)]/X³. From here using the first equation I get (X²-1)(7*3X*5X+5X²)/X³=110X²(X²-1)/X³=110(X²-1)/X. Subsitituing here the values of X and X², and rationalizing the result, brings to the result.
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nice solution
So nice of you.
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GooD 1!
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stay thank
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Very nice sir I am big fan of yours
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Thnku
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x²=(2²-3)=1, plainly.
Sic [7+5-7-5]÷1=0/1
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Thank
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Wow
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It is easier if first show x^3=26-15sqrt(3), and then 1/x^3=26+15sqrt(3).
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Sir than million times
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i lv u 99.9%
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-220root3 be the answer
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Math
相反方程式か
You can say it.....
-220√3
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asnwer=2 isit 😅😅🤣🤣
Please watch the video and compare your answer.
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