A Nice Algebra Problem from Math Olympiad | Can You Simplify?

A Nice Algebra Problem from Math Olympiad | Can You Simplify?
In this video, we'll dive into a fascinating simplification problem that's perfect for anyone preparing for Math Olympiads. Simplification is a crucial skill in algebra, and mastering it can give you a significant edge in competitions. Watch as we break down the steps, explore different techniques, and uncover the beauty of mathematical simplicity. Whether you're a student, a math enthusiast, or preparing for your next Math Olympiad, this challenge is for you!
🔢 What You'll Learn:
Key strategies for simplifying complex expressions
Tips and tricks to approach difficult simplification problems
Step-by-step walkthrough of the solution
🧠 Challenge Yourself:
Pause the video, try to solve the problem on your own, and then watch as we break down the solution. Share your approach and answers in the comments below!
👍 Don't Forget to:
1. Like the video if you found it helpful.
2. Subscribe for more Math Olympiad prep and challenging math problems.
3. Hit the bell icon to get notified of new uploads.
4. Join us and enhance your problem-solving skills. Can you master this simplification problem?
Additional Resources:
• Nice Algebra Challenge...
• Math Olympiad | An Alg...
• Can You Solve This Rad...
• Ready for a Math Chall...
#matholympiad #algebrachallenge #mathsimplification #mathpuzzles #problemsolving #mathtips #mathtutorial #algebra
Thanks for Watching !

[@labbhattacharjee108]

Generalization : y^7+cy^2 =(y^3+ay-b)(*) +(c-2ab)y^2+(b^2-a^3)y+ a^2b so if c=2ab, b^2=a^3 parametrically a=±k^2, b=k^3, c=±2k^5, a^2b=?

[@kassuskassus6263]

E = 2(1152x^2+256x+192)/18x^2+4x+3 =2*64=128

[@mohammedsaysrashid3587]

Thank you Sir 🙏 for sharing....final result 128

[@johnstanley5692]

given g1 = 8*x^3-4*x^2-1 (=0). find 1/x^7+64/x^2 = n/d, n=64*x^5+1, d=x^7. Use synthetic division n/g1 -> 16*x^2+4*x+3,
& d/g1 -> (16*x^2+4*x+3)/128 => n/d = 128 = 1/x^7+64/x^2

[@abcekkdo3749]

128

[@ZhilinChen-my7tp]

利用x的三次方之一等於8減x分之4，兩邊都變成兩次方，很快就有答案了。

[@RealQinnMalloryu4]

x^2+x^2 ➖/ 2+2 ➖ 1+1 ➖/8+8 ➖ =2/16 {x^4/4+2/16} 2x^4/20=10x^4 2^5x^2^2 1^1x^1^2 x^1^2 (x ➖ 2x+1) 1+1 ➖/x^7+x^7 ➖ = 2/x^14 64+64 ➖/x^2+x^2 ➖ 128/x^4 {2/x^14+128/x^4} =130/x^16 =8 .2: 2^3.2^1 1^1.2^1 2 (x ➖ 2x+2)