solve the equation(F4 Math)

(x+2)3次方-1=0

(x + 2)^3 - 1 = 0
(x + 2)^3 - 1^3 = 0
(x + 2 - 1) [(x + 2)^2 + (x + 2) (1) + (1)^2] = 0
(x + 1) {[x^2 + 2(x) (2) + 2^2] + x + 2 + 1} = 0
(x + 1) (x^2 + 4x + 4 + x + 3) = 0
(x + 1) (x^2 + 5x + 7) = 0
x + 1 = 0 or x^2 + 5x + 7 = 0

x + 1 = 0
x = -1

x^2 + 5x + 7 = 0
-------> check b^2 - 4ac
5^2 - 4(1) (7)
= 25-28
<0
NO REAL ROOT

因此 x = -1

我都唔知有沒計錯 .................

我是小學六年級。

(x+2)3次方-1=0
(-1+2)3次方-1=0
(1)3次方-1=0
1-1=0
x=-1

(x+2)3次方-1=0
(x+2)(x+2)(x+2)=0+1
(x+2)/(x+2)/(x+2)=1
1/x+2=1
1=x+2
x=-2+1
x=-1

(x+2)3次方-1=0
(x + 2)^3 = 0 + 1
x + 2 = (1)√3
x + 2 = 1
x = 1 - 2
x = -1