An Amazing Radical Equation | Can You Solve It? | Olympiad

An Amazing Radical Equation | Can You Solve It? | Olympiad
Welcome to infyGyan! In this video, we present an incredible radical equation that will test your algebra skills and challenge your mathematical thinking. This problem is perfect for those preparing for Math Olympiad or anyone who loves tackling tough math problems.
Join us as we break down the steps to solve this intriguing equation, and see if you can find the solution on your own. Don't forget to leave your answers and approach in the comments below!
If you enjoy solving complex math problems, make sure to subscribe to our channel and hit the notification bell for more exciting math challenges.
Additional Resources:
• A Tricky Radical Equat...
• A Nice Radical Equatio...
• Can You Solve This Rad...
• A Nice Factorial Equat...
🔔 Don't forget to subscribe and hit the bell icon to stay updated with our latest math challenges and tutorials. If you enjoy the video, please give it a thumbs up and share it with your friends!
💬 Comment below if you have any questions or if you were able to solve the problem. We love hearing from our viewers!
Thank you for watching and happy problem-solving!

[@dulcineafn]

Where did you conclude a^3+b^3=a^3+b^3+3ab?

[@comdo777]

asnwer=-1/3/2 isit

[@user-kt1dm9jz5t]

X=-1, 2^(1/3), [(1+(11)^(1/2)i)/2]^(1/3), [(1-(11)^(1/2)i)/2]^(1/3)

[@tejpalsingh366]

x= -1; 2^1/3

[@user-kp2rd5qv8g]

Let t=x^3. Then, t^4-2t^3+2t^2-t-6=0 > t=-1 > x=-1, e^(i pi/3), e^(5 pi i/3). Or, t^3-3t^2+5t-6=0 > t=2 > x=2^1/3. Or, t^2-t+3=0 > x = [1/2(1 +/-√11 i)]^1/3. So, if x is real, x = -1, 2^1/3.

[@nitingl4730]

LHS = (-32)^1/5

[@RealQinnMalloryu4]

(x^5 )➖ ( x^15) ➖ 31={x^25 ➖ x^225} ➖ (31)^2= {x^200 ➖ 961}= 761 7^10 61^1 7^2^5 1^1 7^1^2^5^1 1^1^2^1^1 2^1 (x ➖ 2x+1). (x^3)^2 ➖ (1)^2={ x^9 ➖ 1}= x^8 x^2^3 x^2^1 (x ➖ 2x+1).

[@walterwen2975]

An Amazing Radical Equation: ⁵√(x¹⁵ - 31) = x³ - 1, x ϵ R; x =? Let: y = x³; ⁵√(y⁵ - 31) = y - 1, y⁵ - 31 = (y - 1)⁵ = y⁵ - 5y⁴ + 10y³ - 10y² + 5y - 1 5y⁴ - 10y³ + 10y² - 5y - 30 = 5(y⁴ - 2y³ + 2y² - y - 6) = 0, y⁴ - 2y³ + 2y² - y - 6 = 0 (y⁴ - 2y³ + y²) + (y² - y) - 6 = (y² - y)² + (y² - y) - 6 = (y² - y - 2)(y² - y + 3) = 0 (y - 2)(y + 1)(y² - y + 3) = 0, y² - y + 3 = 0, y = (1 ± i√11)/2; Rejected, y = x³ ϵ R y - 2 = 0, y = x³ = 2; x = ³√2 or y + 1 = 0, y = x³ = - 1; x = - 1 Answer check: x = ³√2 or x = - 1 ⁵√(x¹⁵ - 31) = ⁵√[(³√2)¹⁵ - 31] = ⁵√(2⁵ - 31) = 1, x³ - 1 = (³√2)³ - 1 = 1; Confirmed ⁵√(- 1 - 31) = ⁵√(- 32) = - 2, x³ - 1 = (- 1)³ - 1 = - 2; Confirmed Final answer: x = ³√2 or x = - 1