Solving A Nice Radical

[@erikroberts8307]

Excellent!!!

[@DrQuatsch]

I used the "first method", I just made a short cut after cubing, as I noticed 4x + 4 is just 4 * (x + 1), so after squaring I got 16 * (x + 1)^2 = (x^2 + 8x + 16)(x + 1). Put everything on the same side and you already have your (x + 1) factor and the 16s cancel. So you are left with (x + 1)(x^2 - 8x) = 0 or x(x + 1)(x - 8) = 0.

[@dan-florinchereches4892]

We can observe X=0 and X=-1 verify the original equation.
If X=y^3 this corresponds 1-1 on the domain and we get
√(y^3+1)-y=1
√(y^3+1)=1+y square everything
y^3+1=y^2+2y+1
y(y^2-y-2)=0
y(y+1)(y-2)=0
So we can have y€{-1,0,2}
So x€{-1,0,8}
Checking X=8 to verify
√9-8^(1/3)=3-2=1 is a solution

[@wongmanwaihehe]

Subbing in 0 is the first thing in my mind

[@Quest3669]

Lets x= a^3 to easily drove through 3 answrs

[@فیروزاهنگری]

x+1=t^2 ,t-1=(t^2-1)^1/3 ,(t-1)^3=t^2--1 ,t(t-3)(t-1)=0 ,t=0,t=3,t=1. then. X=-1. ,8. ,0

[@prollysine]

let u=x^(1/3) , x=u^3 , u^3+1=1+2u+u^2 , u^3-u^2-2u=0 , u(u^2-u-2)=0 , u=0 , u^3=0 , x=0 , u^2-u-2 , u=(1+/-V(1+8))/2 ,
u= (1+3)/2 , (1-3)/2 , u= 2 , -1 , x=u^3 , x= 0 , 8 , -1 , test , x=0 , V(0+1)-0=1 , OK ,
test , x=8 , V9-(8)^(1/3)=3-2 , --> 1 , OK , test , x= -1 , V(-1+1)-(-1)^(1/3)=0+1 , --> 1 , OK , solu , x= 0 , 8 , -1 ,

[@mega_mango]

x^1/3 = n
n³+1 = (n+1)²
(n+1)(n²-2n) = (n+1)(n-2)n = 0
n = {-1,0,2}
x = n³ = {-1,0,8}

[@robertveith6383]

The first line is wrong. The fractional exoonent must be inside grouping symbols: x^(1/3).

[@mega_mango]

@@robertveith6383 cry about it