A Math Olympiad Expression | You Should Try This!

A Math Olympiad Expression | You Should Try This!
Dive into the world of Math Olympiad with this exhilarating challenge! Can you simplify this complex expression? Put your algebra skills to the test and see if you can crack the code. Join us for an exciting journey of problem-solving and mathematical discovery! 💡🧮 #matholympiad #algebra #problemsolving #mathematics #challengeaccepted #education #stemeducation #criticalthinking #maths #simplification #expression
📘 Topics covered:
Introduction to the Three Methods
Expressions
Math Olympiad
Math Olympiad Preparation
Rational Expression
Algebraic Identities
Algebra
Pascal's triangle
Exponents
Exponent laws
Substitutions
Binomial expansion
Math Tutorial
Problem solving
Math Enthusiasts
Mathematics
Math skills
Timestamps:
0:00 Introduction
0:26 Method-1
0:30 Properties of exponents
2:15 Substitutions
4:00 Binomial expansion
7:15 Algebraic identities
9:30 Method-2
11:20 Method-3
13:24 Answer
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We'd love to hear from you! Did you manage to simplify the expression? What other math problems would you like us to cover? Let us know in the comments below!
🎓 Happy learning, and see you in the next video! 🎉
Thanks for Watching !!
@infyGyan

[@mohammedsaysrashid3587]

Thanks...(M-3 ) was a wonderful one's...final result 9,99,001....thanks for sharing

[@mikelivstone]

Sometimes these problems are more easily solved from the middle. You be the judge:
Let a = 500, b=499, and note that 1= 1^1 = 1^2 = 1^4. Then, from 2:09:
E = [(a-b)^4 + (a+b)^4 + (2a)^4] / [(a-b)^2 + (a+b)^2 + (2a)^2]
= [(a^4 - 4ba^3 + 6(ab)^2 - 4ab^3 + b^4) + (a^4 + 4ba^3 + 6(ab)^2 + 4ab^3 + b^4) +16a^4] / [(a^2 -2ab +b^2) + (a^2 +2ab +b^2) + 4a^2]
= (18a^4 + 12(ab)^2 + 2b^4) / (6a^2 + 2b^2)
= (2/2) * (3a^2 + b^2)^2/(3a^2 + b^2)
= 3a^2 + b^2
Since b = a-1:
E = 3a^2 + (a-1)^2 = 3a^2 + a^2 -2a + 1 = 4a^2 - 2a + 1 = (2a)^2 -(2a) + 1 = 1000^2 -1000 + 1 = 999001

[@infyGyan]

@mikelivstone
It's great! Thanks for sharing an interesting way of solving such problems.
@infyGyan

[@abcekkdo3749]

999001