Are You Smart Enough To Crack This Algebra Challenge?

Are You Smart Enough To Crack This Algebra Challenge?

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In this video, we explore an amazing algebra problem. If you love solving complex algebraic 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:

Algebra Challenge
Algebraic manipulations
Problem-solving techniques
Reciprocal equation
Binomial expansion
Synthetic division method
Rational root theorem
Square root
Algebraic identities
College Math

🔢 What You'll Learn:

Key strategies for simplifying cube root and square root in radicals
Tips and tricks to approach difficult radical 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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[ManojkantSamal]

be
^=read as to the power
*=read as square root
Let a=x, b=(1/x)
ab=x(1/x)=1....eqn1
a^5+b^5=(a+b)^5 - [5ab(a+b){(a+b)^2-ab}]
Let R=a+b
So,
a^5+b^5=R^5-{5R(R^2-1)}
a^5+b^5=R^5- 5R^3+5R
As per question
a^5+b^5+5=9^3
a^5+b^5=729-5
a^5+b^5=724
So,
724=R^5 - 5R^3 +5R
Though 5th powered equation, So application of H &T method come into action to find out the first root
So,
R^5 - 5R^3 +5R
Let R=4
(4^5) -{5(4^3)}+(5×4)
=1024-(5×64)+20
=1024-320+20
=1024-300=724
So, R=4
a+b=4
(a+b)^=a^3+b^3+3ab(a+b)
(4^3)=a^3+b^3+(3×4)
64=a^3+b^3+12
a^3+b^3=64-12=52
a^3+b^3-3=52-3
X^3+(1/x^3)-3=49
As per question
X^3+(1/x^3)-3=a^2
So,
a^2=49
a=*(49)=±7

[RealQinnMalloryu]

{x^5+x^5 ➖ }+{5+5 ➖ }+{1+1 ➖ }/{x^5+x^5 ➖ 2^6x^1 2^2^3x^1 1^2^3x^1 2^3x (x ➖ 3x+2) (x^3)^2 ➖ (3)^2 {1+1 ➖ }/{x^3+x^3 ➖ }={x^9➖ 9}2/x^6={x^0+x^0 ➖ x^0+x^0 ➖ ➖ 3x+2).

[ShriH-do]

Let k= x+1/x; =>
k^5={asx*1/x=1}=
x^5+(1/x)^5+
5{(k^3-3k)}+10(k)=> k^5= 724+5k^3 -5k ;=>
k^5-5k^3+5k-724 =0; => => => k=4;
x^3+(1/x^3)-3=
k^3-3k-3=a^2; for
k=4, a^2= 49; =>
a = ±7
{Note: k(k^4-5k^2+5)= 724= 4.181 =>
k=4; verified for k^4-5k^2+5=181}