A Nice Radical Equation | Math Olympiad | Algebra Challenge 

I'm with you up to 1:20, but it's really simple at that point just to cube both sides and solve for a: (a^6 - 28)^(1/3) = a^2 - 4 a^6 - 28 = a^6 -12a^4 + 48a^2 - 64 Bring everything to the left-hand side. The a^6 terms drop out, leaving a biquadratic equation: (It's probably also worth mentioning that this is the step that explains why what appears from the outset to be an expression of the sixth degree has only 4 solutions.) 12a^4 - 48a^2 + 36 = 0 a^4 - 4a^2 + 3 = 0 (a^2 - 3)(a^2 - 1) = 0 a = ±1 or ±sqrt3 x = a - 2 = -1, -3, or -2 ± sqrt3

Let a^3=(x+2) ^6-28; b=(x+2) ^2;x^2+4x=b-4; b^2-a^3=28, ab=-3; a^2+4a+3=0 x=-1, -3, -2±√3.

Η εξισωση γινεται (y+4)^3-y^3=28 οπου y=χ^2+4χ τελικα χ=-1, -3, -2+-(3)^1/2

Если sqrt (3)(a^6-28)=a^2-4 Можно все просто возвести в куб и получить биквадратное уравнение (квадратное относительно а^2). Решение не составляет труда

X=-1, -3, (3)^(1/2)-2, -(3)^(1/2)-2.

Let x+2=t and a=t^2. Then, the given equation reads a^3-28=(a-4)^3 > a^2-4a+3=0 > a=3,1. a=3 > x=-2 +/-√3 and a=1 > x= -1,-3. All four are valid solutions. So, x= -3, -1, -2 +/-√3.

X=-1,-3,√3 -2, -√3 -2

At 1:38, how about letting a² = b then you get 12b² - 48b +36 = 0 b² - 4b +3 = 0 (b-3)(b-1) = 0 b= 1, 3 a = ± 1, ± √3 x = -2 ± 1, -2 ± √3 x = -1, -3, -2 + √3, 2 - √3 and then verify

We see that (x+2)^2=x^2+4x+4 and x*(x+4)=x^2+4x. Both have common the term x^2+4x. Let x^2+4x=t . So x^2+4x+4=t+4 and (x+2)^6=(t+4)^3 . Then rising BOth Sides of EQUATION(1) to 6th we get (t+4)^3-28=t^3 so t^3+3*t^2*4+3*t*4^2+4^3-28-t^3=0 so 12*t^2+48t+36=0 so 12*(t^2+4t+3)=0 so 12*(t+1)*(t+3)=0 and 12>0 so t=-1 OR t=-3. CASE 1: t=-1 so x^2+4x=-1 so x^2+4x+1=0. D=4^2-4*1*1=16-4=12 so x=(-4+ -sqrt12) /2=(-4+ -2*sqrt3)/2= -2+ -sqrt3** CASE 2 : t=-3 so x^2+4x=-3 so x^2+4x+3=0 so (x+1)*(x+3)=0 so x=-1 or x=-3. Finally we have 4 solutions for x={ -2-sqrt3, -3, -1, -2+sqrt3} in ascending order.

Write the root index in the correct location.

X= -1; -3; +-√3-2

{x^3+32x^6} ➖ 28=32x^9 (28)^2=784{32x^9 ➖ 784}=752x^9 10^70^52x^9 10^7^102^26x^9 2^57^1^2^5^2^13x^9 1^1^1^1^1^12^13^1x^3^2 2^1^1x^3^1 2x^3 (x ➖ 3x+2). {x^2+2x}= 2x^3 (x ➖ 3x+2).

X=-1

∛[(x + 2)⁶ - 28] = x(x + 4) find x ∈ ℝ x + 2 = u => x = u - 2 ∛(u⁶ - 28) = (u - 2)(u + 2) ∛(u⁶ - 28) = u² - 4 u⁶ - 28 = v³ => u⁶ - v³ = 28 v = u² - 4 => u² - v = 4 u⁶ - v³ = 7(u² - v) (u² - v)(u⁴ + u²v + v² - 7) = 0 u² - v = 4 ≠ 0 u⁴ + u²v + v² - 7 = 0 u² = v + 4 (v + 4)² + v( v + 4) + v² - 7 = 0 v² + 8v + 16 + v² + 4v + v² - 7 = 0 3v² + 12v + 9 = 0 v² + 4v + 3 = 0 (v + 1)(v + 3) = 0 v = - 1 => u² - 4 = -1 => u² = 3 => u = ± √3 v = - 3 => u² - 4 = -3 => u² = 1 => u = ± 1 x = u - 2 *x = {-1, -3, -2 + √3, -2 - √3}*