Math Olympiad | A Nice Rational Equation | 90% Failed to solve

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[@rorydaulton6858]

I solved it differently. In the initial equation, let u = 1/x. Then
u⁵ - 2u⁴ + 4u³ - 8u² + 16u - 32 = 0.
Seeing the pattern of powers of 2 gives us the idea to multiply both sides of that equation by u + 2, giving us (after much cancellation of terms),
u⁶ - 64 = 0
u⁶ = 2⁶
(1/x)⁶ = 2⁶
x⁶ = (½)⁶
In the complex numbers, this can only happen when
x = ε·½ where ε is a 6th root of 1. Those roots are
ε = e^(n·2πi/6) for n = 0, 1, 2, 3, 4, 5. Our multiplication by u + 2 gave us an extraneous root of u = -2 (x = -½) so we remove n = 3. Using Euler's formula
e^(n·2πi/6) = cos(n·2π/6) + sin(n·2π/6)i
and multiplying each by ½ gives all five of the answers.

[@vijaymaths5483]

Good one...Thank you for sharing! 😊

[@TrinityRed]

@@vijaymaths5483 u=1/x-2 is much easier

[@RealQinnMalloryu4]

(1)^2/(x^5)^2= 1/x^25(2)^2/(x^4)^2=4/x^16 {1/x^25 ➖ 4/x^16} =3/x^9 {4x+4x ➖ }/{x^3+x^3 ➖ }=8x^2 /x^6 {3/x^9+8x^2/x^6}=11x^2/x^15 (8)^2/(x^2)^2=64/x^4 {11x^2/x^15 ➖ 64/x^4}=53x^2/x^11 {16x+16x ➖ }/{x+x ➖ } =32x^2/x^2 /{53x^2/x^11+32x^2/x^2}={85x^4/x^13}=85x^3.1 5^17x^3^1 5^17^1x^3^1 5^1^1x^3^1 5x^3^1 5^1 x^3^1 1^1x^3^1 x^3^1" (x ➖ 3x+1).

[@TrinityRed]

Your solution is too complicated u=1/x-2 (2 minutes) and after exponential (cos and sin).