Olympiad Math hexic Equation |Tricks of Solving (x+1)^6=64 | Solving Olympiad Algebra Challenge

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In this Math Olympiad Algebra Problem, you'll learn tips and tricks of solving International Math Olympiad exams quickly. #IMO #matholympiad #algebra #radicalequations #simplify #exponential

[@mircoceccarelli6689]

👍👍👍

[@anderlecht1969]

Nice one. You are a good teacher.

[@superacademy247]

Thank you! 😃

[@sambhavkhandelwal6470]

simple method: (x+1)^6 = 2^6 When the exponent is equal, the bases are equal x+1 = 2 and x+1 = -2 x = 1 and -3

[@yuusufliibaan1380]

❤❤❤ thanks 💯💯💯

[@superacademy247]

Welcome 😊

[@mehmethancicek3372]

Thanks

[@superacademy247]

Welcome

[@dasliebesgluckprinzip9635]

2^6=64 X+1=2 Rest is simpel

[@user-kp2rd5qv8g]

Let z = (x+1)/2. Then, z^6 = 1. So, z = e^(2 pi i n/6) = e^(i pi n/3), n = 0,1,2,3,4,5. So, z = 1, 1/2[ 1 + sqrt(3)i], 1/2[ -1 + sqrt(3)i], -1, - 1/2[ 1 + sqrt(3)i], 1/2[ 1 - sqrt(3)i]. Thus, x = 2z-1 = 1, sqrt(3) i, -2 + sqrt(3)i, -3, -2 - sqrt(3) i, -sqrt(3)i.