(3x+2)^1/3 - (x-1)^1/3 = 1

Let us cube both sides:

[(3x+2)^1/3) - (x-1)^1/3]^3 = 1

But we know that:

(a-b)^3 = a^3 - b^3 - 3ab(a-b)

==> [(3x+2)^1/3 - (x-1)^1/3]^3 = (3x+2) - (x-1) - 3(3x+2)^1/3*(x-1)^1/3 * [(3x+2)^1/3 - (x-1)^1/3] = 1

==> 2x +2 - 3[(3x+2)(x-1)]^1/3*(1)=1

==> -3[(3x+2)(x-1)]^1/3 = -2x-2

==> 3(3x+2)(x-1)]^1/3 = 2x+2

Cube both sides:

==> 27(3x+2)(x-1) = (2x+2)^3

==> 27(3x^2 -x -2) = 8x^3 + 24x^2 + 24x + 8

==> 81x^2 - 27x - 54= 8x^3 + 24 x^2 + 24x + 8

Group similar:

==> 8x^3 -57x^2 + 51x+ 62=0

==> (x-2)(8x^2-41x-31) = 0

x1= 2

x2= [41+ sqrt(1681 +992)]/16 = (41+51.7)/16 = 5.8

x2= (41- 51.7)/16 = -0.67

Because of the cube root, the constraints of existence are no longer necessary.

We'll solve the equation using the formula:

(a - b)^3=a^3 - b^3-3ab(a-b)

We'll raise to cube both sides:

[(3x+2)^1/3 - (x-1)^1/3]^3 = 1

(3x+2) - (x-1)-3[(3x+2)*(x-1)]^1/3*[(3x+2)^1/3 - (x-1)^1/3]=1

But, from enunciation, [(3x+2)^1/3 - (x-1)^1/3] = 1.

The expression will become:

3x+2-x+1-3[(3x+2)*(x-1)]^1/3*1 = 1

We'll combine like terms:

2x+2 = 3[(3x+2)*(x-1)]^1/3

We'll factorize:

2(x+1) = 3[(3x+2)*(x-1)]^1/3

We'll raise to cube again:

[2(x+1)]^3 =[3[(3x+2)*(x-1)]^1/3]^3

8(x^3 + 3x^2 + 3x + 1) = 27(3x^2 - 3x +2x - 2)

We'll remove the brackets and we'll combine like terms:

8x^3 - 57x^2 + 51x + 62 = 0

(x-2)(8x^2-41x-31)=0

x-2 = 0

x=2

8x^2-41x-31 = 0

We'll apply the quadratic formula:

x1 =(41+sqrt2673)/16

x1 = (41+9sqrt33)/16

x2 = (41-9sqrt33)/16

**The solutions are: {2 ; (41+9sqrt33)/16 ; (41-9sqrt33)/16}.**