【問題】請教一題高次代數

[curry 10]

[x+1][x二次方+1][x三次方+1]=30*x三次方

[chpohoa1 10]

顯然x=0不為其解

兩邊同除以x^3

=> [√x + 1/√x] * [x + 1/x] * [√(x^3) + 1/(√x^3)] = 30

令 √x + 1/√x = t

x + 1/x = t^2 - 2

√(x^3) + 1/(√x^3) = t^3 - 3t

=> t^2 * ( t^2 - 2 ) * (t^2-3) = 30

令 t^2 = u => (u^2 + 6)(u-5) = 0

=>x + 1/x = t^2 - 2 = -2 ± √6i ，3

=> x^2 + ( 2∓√6i) x + 1 = 0 , x^2 - 3x + 1 = 0