A Nice System of Radical Equations | Math Olympiad Preparation

A Nice System of Radical Equations | Math Olympiad Preparation
Welcome to our Math Olympiad Prep series! In this video, we'll tackle an amazing system of equations that's sure to challenge and sharpen your problem-solving skills. Perfect for those preparing for the Math Olympiad or anyone looking to enhance their math prowess, this tutorial will guide you step-by-step through the solution process. Don't miss out on this exciting opportunity to master a complex system of equations for the competition!
Stay tuned, and let’s solve this together! If you enjoy the video, please like, subscribe, and hit the bell icon for more Math Olympiad tips and tricks.
Topics covered:
System of equations
Algebra
Problem solving
Algebraic identities
Algebraic manipulations
Solving systems of equations
Factorization
Cubic equations
Math enthusiast
Math tutorial
Math Olympiad
Math Olympiad Preparation
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Thanks for watching!

[@Shobhamaths]

Let a=³√x, b=³√y,
b=3+a, b^5=a^5+33,
Substitute b=3+a then
5a^4+30a^3+90a^2+135a+70=0;
a=-1, -2;
(x, y) =(-1, 8),( -8, 1) 👍

[@user-kp2rd5qv8g]

Let y^1/3=a, x^1/3=b, ab=t. Then, a-b=3, a^5-b^5=33. Using (a-b)^5 = a^5-b^5 -5ab[(a-b)2+3ab] +10a^2b^2(a-b), we get (t+2)(t+7)=0. t =-7 does not yield real a,b. t=-2 gives (a,b) = (1,-2), (2,-1) > (x,y) = (-8,1), (-1,8).

[@ExquisiteHappiness]

(x,y) = (-1,8) or (-8,1) for real x and y

[@ΜαργαριταΚαναρη]

Αν y^(1/3)=α και χ^(1/3)=β εχω το συστημα α-β=3 και α^5- β^5=33 τελικα ( α,β)=(2,-1) ,(1,-2) .αρα (y,χ)=(8,-1) , (1,-8)

[@user-kt1dm9jz5t]

X,Y=（-8, 1）; （-1, 8）.

[@SidneiMV]

3 + ∛x = ∛y
33 + x∛x² = y∛y²
∛x = -a
∛y = b
3 - a = b
33 - a⁵ = b⁵
a + b = 3
a⁵ + b⁵ = 33
a⁴ - a³b + a²b² - ab³ + b⁴ = 11
a⁴ + b⁴ - ab(a² + b²) + (ab)² = 11
(a + b)² = 9 = a² + b² + 2ab
a² + b² = 9 - 2ab
(a² + b²)² = (9 - 2ab)² = a⁴ + b⁴ + 2(ab)²
a⁴ + b⁴ = (9 - 2ab)² - 2(ab)²
a⁴ + b⁴ = 2(ab)² - 36ab + 81
2(ab)² - 36ab + 81 - ab(9 - 2ab) + (ab)² = 11
5(ab)² - 45ab + 70 = 0
(ab)² - 9ab + 14 = 0
(ab - 7)(ab - 2) = 0
ab = 7
a + b = 3
t² - 3t + 7 = 0 => complex roots
ab = 2
a + b = 3
(a, b) = {(1, 2); (2, 1)}
a = 1 => ∛x = -1 => *x = -1*
b = 2 => ∛y = 2 => *y = 8*
a = 2 => ∛x = -2 => *x = -8*
b = 1 => ∛y = 1 => *y = 1*