An Amazing Algebra Challenge | 95% Failed To Solve! | Give It A Try!

An Amazing Algebra Challenge | 95% Failed To Solve! | Give It A Try!

Welcome to infyGyan!

In today's video, we explore a fascinating algebraic simplification problem. If you love solving complex expressions and pushing your problem-solving skills to the limit, this is the perfect challenge for you. Join me as we work through this intricate algebra problem step-by-step. Don't forget to like, comment, and subscribe for more math challenges and solutions!

📌 Topics Covered:

Algebraic simplification
Algebraic manipulations
Problem-solving techniques
Algebraic identities
Math Olympiad problems

🔢 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!

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Thank You for Watching videos!!

[parasuramankarmegam]

Excellent way of teaching step by step

[RashmiRay-cy]

Note that (a^2+b^2+c^2)(a^5+b^5+c^5) = a^7+b^7+c^7 +3abc(a^2b^2+b^2c^2+c^2a^2) -a^2b^2c^2(a+b+c) = a^7+b^7+c^7 +3abc(a^2b^2+b^2c^2+c^2a^2). So, E = But using a^2+b^2+c^2 = -2(ab+bc+ca) and a^3+b^3+c^3 = 3abc, we can show that a^5+b^5+c^5 = -5abc(ab+bc+ca). So, (a^2+b^2+c^2)(a^5+b^5+c^5)= -10abc(ab+bc+ca)^2. But as a+b+c=0, (ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2. Putting everything together, E = 1 -3/10 = 7/10.

[johnlv]

Very nice problem and solution.

Conjecture:
Suppose
i) n is a positive integer>3
ii) m is a positive integer with n/2 <= m <= n-2
iii) a+b+c=0
iv) is not 0.

Then,
is a constant in only these cases:
n=4, n=5, or n=7.
All m in the range n/2 <= m <= n-2 for these three values of n make the ratio constant.
In other words, the given problem (the case n=7 and m=5) is one of only 4 cases where it magically works.

[RealQinnMalloryu]

{a^7a^7 ➖}+{b^7+b^7 ➖ }+{c^7+c^7 (abc ➖ 1abc+1).
.

[sendai-shimin]

c = -(a + b)
a^2 + b^2 + c^2 = 2(a^2 + b^2 + ab)
a^5 + b^5 + c^5 = -5(a^4b + 2a^3b^2 + 2a^2b^3 + ab^4)
(a^2 + b^2 + c^2)(a^5 + b^5 + c^5) = -10(a^6b + 3a^5b^2 + 5a^4b^3 + 5a^3b^4 + 3a^2b^5 + ab^6)
and
a^7 + b^7 + c^7 = -7(a^6b + 3a^5b^2 + 5a^4b^3 + 5a^3b^4 + 3a^2b^5 + ab^6)

[johnlv]

The question should have stated that a, b, c must be restricted so that the denominator is not 0.

[gregevgeni]

7/10
With some theory.
Let x^3 + K x^2 + L x + M = 0 the polynomial with roots a, b, c .
Then x^n ( x^3 + K x^2 +L x + M) = 0
x^(n+3) + K x^(n+2) + L x^(n+1) +
+ M x^n = 0 (*)
For x = a, b, c we have from (*)
a^(n+3) + K a^(n+2) + K a^(n+1) +
+ M a^n = 0 (1)
Accordingly
b^(n+3) + K b^(n+2) + L b^(n+1) +
+ M b^n = 0 (2).
c^(n+3) + K c^(n+2) + K c^(n+1) +
M c^n = 0 (3)
Summing the (1), (2), (3)
(a^(n+3) + b^(n+3) + c^(n+3)) +
K (a^(n+2) + b^(n +2) + c^(n+2)) +
L (a^(n+1)+ b^(n+1) +c^(n+1)) +
M (a^n + b^n + c^n) = 0. (**).
Let S_n = a^n + b^n + c^n, then
S_(n+1) = a^(n+1) +b^(n+1) + c^(n+1),
S_(n+2) = a^(n+2) + b^(n+2) + c^(n+2) and
S_(n+3) = a^(n+3) + b^(n+3) + c^(n+3) (***) . The (**) due to (***)
written
S_(n+3) + K S_(n+2) + L S_(n+1) +
+ M S_n = 0 (#)
For the polynomial
x^3 + K x^2 + L x + M = 0
we have that
a + b + c = - K, ab + bc + ca = L and
abc = - M ( Vieta's formulae).
Too, from hypothesis of the given problem it's a+ b +c = 0, thus K = 0.
So, with some algebra,
S_1 = a + b + c = - K = 0
S_2 = a^2 + b^2 +c^2 =
= (a +b+c)^2 -2(ab +bc + ca) =>
S_2 = - 2 L,
S_3 = a^3 + b^3 +c^3 = 3abc =>
S_3 = - 3 M ( Euler's Identity).
S_4 = - L S_2 - M S_1 = L^2,
S_5 = ... = 5 LM,
S_6 = ... = -2 L^3 + 3 M^2
S_7 = ... = -7 L^2 M
. ... ..
Now we come back to the given problem :
S_7 / S_2 • S_5 =
( -7 L^2 M) / (- 2 L)( 5 LM) = 7/10.

[Fjfurufjdfjd]

Ενας αλλος τροπος αν δεν παει το μυαλο μας στην ταυτοτητα που χρησιμοποιησατε. Ισχυει α^3+β^3+γ^3=3αβγ (1) α^2+β^2+ γ^2=-2(αβ+βγ+γα) (2) (1)×(2) (3) Π(παρονομαστης) αλλα (επειδη α+β+γ=0). ( αριθμητης) τελικα