Solve equation

If the question is x^3 + (1 + 2^1.5)x^2 + (1 + 2^1.5x) + 1 = 0, then the solution is simple.
(x + 1)(x^2 + 2^1.5x + 1) = 0
x = -1 or x = [2^1.5 +/- √(2^3 - 4)]/2
x = -1 or x = √2 +/- 1
If the question is correct, then there is no simple analytic method to obtain the root of the equation x^3 + (1 + √8)x^2 + (1 + √8)x – 1 = 0
Now let y = x^3 + (1 + √8)x^2 + (1 + √8)x – 1
First determine the number of real roots for the equation.
dy/dx = 3x^2 + 2(1 + √8)x + (1 + √8) = 0
x = {-2(1 + √8) +/- √[4(1 + √8)^2 - 12(1 + √8)]}/6
x = -0.68 & -1.87
d^2y/dx^2 = 6x + 2(1 + √8)
when x = -0.68; d^2y/dx^2 = 3.56 => min and min = -2.15
when x = -1.87; d^2y/dx^2 = -3.56 => max and max = -1.3
Both local max and min are smaller than zero.
From these results, it can be concluded that y = x^3 + (1 + √8)x^2 + (1 + √8)x – 1 = 0 has only one real root.
y(0) = -1 < 0
y(1) = 2 + 2√8 > 0
Therefore the only root lies between x = 0 and x = 1
Using Newton’s method (or other numerical method), the root is found to be 0.2132
Newton’s Method :
x(n+1) = x(n) – f[x(n)]/f’[x(n)]
Arbitrarily take x(0) = 1
x(1) = 1 – 7.657/14.49 = 0.4714
x(2) = 0.4714 – 1.760/8.105 = 0.2542
x(3) = 0.2542 – 0.2371/5.969 = 0.2145
x(4) = 0.2145 – 0.007/5.609 = 0.2132
x(5) = 0.2132 – 0.000/5.597 = 0.2132
The same result can be obtained by direct calculation using the cubic root formula available in wikipedia:
http://en.wikipedia.org/wiki/Cubic_equation
Also try this site for cubic root calculation online:
http://www.easycalculation.com/algebra/cubic-equation.php

I think the equation is x^3 + (1+2^1.5)x^2 + (1+2^1.5)x + 1 = 0

代A=1+2^1.5,
x^3+Ax^2+Ax-1=0
x (x^2+Ax+A) =1
x=1 or x^2+Ax+A =1
x^2+Ax+A-1=0
(x+1)(x+u)=0
x=-1 or x=-A

so x=1,x=-1 , x=-(1+2^1.5)