x^3+x=1 ,so x=?

ax^3+bx^2+cx+d = 0 x^3 + x -1 = 0 where a = 1, b = 0 c = 1, d = -1By Descarte’s rule, 有一個real root 和兩個unreal roots.
方法1: Cardano 公式x = {q + [q^2 + (r-p^2)^3]^1/2}^1/3 + {q - [q^2 + (r-p^2)^3]^1/2}^1/3 + pwhere p = -b/(3a), q = p^3 + (bc-3ad)/(6a^2), r = c/(3a)p = 0, q = + (-3(-1))/(6) = 1/2, r = 1/3x = {q + [q^2 + (r-p^2)^3]^1/2}^1/3 + {q - [q^2 + (r-p^2)^3]^1/2}^1/3 + px = {1/2 + [1/4 +1/27]^1/2}^1/3 + {1/2 - [1/4 + 1/27]^1/2}^1/3 x = {1/2 + [31/108]^1/2}^1/3 + {1/2 - [31/108]^1/2}^1/3 x = {0.5 + 0.535758375}^1/3 + {0.5 - 0.535758375}^1/3 x = 1.011780142 – 0.329452336x = 0.6823278038
方法2: By Vieta’s substitution 請查閱工程數學教科書
方法3: Newton’s method (微積分)x(n+1) = xn - f(x) / f '(x) where f '(x) is the derivative.xn = x - (x^3 + + x - 1)/(3x^2 + 1) Trial 1: put x = 0.68, xn = 0.68 - (0.68^3 + + 0.68 -1)/(3(0.68)^2 + 1) = 0.6823434Trial 2: put x = 0.6823434, xn = 0.6823278038
方法4: 用電腦寫計算機編程(我用Fortran語言, 因有complex number function)x = 0.6823277x = -0.3411639 + 1.161541i, x = -0.3411639 - 1.161541i
方法5: 用 Graphic Calculator, Check:x = 0.6823278038x^3 + x -1 = 00.6823278038^3 + 0.6823278038 -1 = - 0.00000000006

幾好,不過無step+通解(我要人手都計到果d)

http://www.wolframalpha.com/input/?i=x^3%2Bx%3D1
-> exact solution

x^3+x=1，So x=?
解l
x^3+x－1=0
(x^3－0.6824x^2)+(0.6824x^2－0.46567x)+(1.46567x－1)=0
x^2(x－0.6824)+0.6824x(x－0.6824)+1.46567(x－0.6824)=0
(x^2+0.6824x+1.46567)(x－0.6824)=0
x^2+0.6824x+1.46567=0
x=[－0.6824+/-√(0.46567－4*1.46567)]/2
=－0.3412+/-1.16157i
x=0.6824或 x=－0.3412+/-1.16157I