Poland Math Olympiad | A Nice Radical Challenge | Algebra

Poland Math Olympiad | A Nice Radical Challenge | Algebra
Welcome to another exciting Math Olympiad challenge! In this video, we solve a fascinating radical problem from the Poland Math Olympiad. Join us as we delve into this intriguing algebraic challenge and explore the elegant solutions. Whether you're a math enthusiast, a student preparing for competitions, or simply love solving problems, this video is for you. Let's solve this together and enhance our problem-solving skills!
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🔍 In this video:
Detailed walkthrough of a challenging algebra problem from the Math Olympiad.
Tips and tricks for solving complex algebraic equations.
Encouragement to enhance your problem-solving skills and mathematical thinking.
📣 Call to Action:
Have a go at the problem yourself before watching the solution!
Share your solutions and approaches in the comments below.
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#matholympiad #poland #algebra #radical #mathproblems #maths #mathcompetition #problemsolving #learnmaths #mathematics
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Thank You for Watching!!

[@mohammedsaysrashid3587]

Thank you for sharing...Sir 🙏....E=3

[@user-kp2rd5qv8g]

Let a=3^1/5 and b = 2^1/5. Then x = a^4+a^3b+a^2b^2+ab^3+b^4 = (a^5-b^5)/(a-b) = 1/(a-b) > 1/x = a-b = 3^1/5-2^1/5 > 1/x + 2^1/5 = 3^1/5 > (1/x + 2^1/5 )^5 = 3.

[@abcekkdo3749]

E=3

[@user-ny6jf9is3t]

Χρησιμοποιω την ταυτοτητα (β^5-α^5)/(β-α) =β^4+β^3×α+β^2×α^2+β×α^3+α^4 οπου β=3^(1/5) και α=2^(1/🎉5). Εχω τελικα Ε=(3^1/5-2^1/5+2^1/5)^5=(3^1/5)^5=3

[@RealQinnMalloryu4]

{405+270}="675 {180+120} =300 {675+300}= 975 {975+90} =1065 100^100 5^13 10^10^10^10 5^13^1 2^52^52^52^5 5^1^1 1^1^1^11^12^1 5^1 2^1 1^1 21 (x ➖ 2x+1). {2/x^2+ 32} =34/x^2 =17 8^9 2^33^2 1^1^2^1^1 2^1(x ➖ 2x+1).