(7x + 1) ^1/3 - x = 1

First we ill move x to the left side:

==> (7x+1)^1/3 = x+ 1

Now cube both sides:

==? [(7x+1)^1/3]^3 = (x+1)^3

==> (7x + 1) == x^3 + 3x^2 + 3x + 1

Now combine like terms:

==> x^3 + 3x^2 + 3x - 7x + 1 - 1 = 0

==> x^3 + 3x^2 - 4x = 0

Now let us factor x:

==> x(x^2 + 3x - 4) = 0

Now factor between brackets:

==> x( x+4)(x-1) = 0

==> x1= 0

==> x2= -4

==> x3= 1

Then we have 3 possible x values:

**x = { 0, 1, -4}**

To find x if (7x+1)^(1/3) - x = 1.

Substitute 7x+1 = t^3, x = (t^3-1)/7 in the given equation :

t - (t^3-1)/7 = 1

7t -t^3 +1 = 7

7t -t^3 +1- 7 = 0

7t-t^3 -6 = 0

Multiply by -1:

t^3 -7t + 6 = 0 which is satisfies for t =1 as 1^3-7*1+6 = 0.

Therefore , t-1 factor.

Similarly if t= 2 also satisfy t^2-7t+6 = 0, as 2^3-2*7+6 = 0.

Also for t = -3, t^3-7t+6 = (-3)^3-7(-3)+6 = -27+21+6 = 0.

t =1 gives x = (t^3-1)/7 = (1-1)/7 = 0

t = 2 gives x = (2^3-1)/7 = 1

t = -3 gives x = (-3^3-1)/7 = -4.

Thereore x = 0, 1 and -4.

Verification: Put x= 0 in (7x+1)^(1/3) -x = (7*0+1)^(1/3)-0 =1-0 = 1, verifies with RHS.

Put x = 1 in (7x+1)^(1/3) - x = (7*1+1)^(1/3)-1 = 2-1 = 1 verifies RHS.

Put x = -4 in (7x+1)^(1/3) - x = (7*(-4) +1)^(1/3) - (-4) = (-27)^(1/3) - (-4) = -3+4 = 1 verifies with RHS.

Since the radical is of 3rd order, we don't have to impose constraints of existence of the radical.

The first step is to add x both sides:

(7x + 1)^1/3 = 1+ x

The next step is to raise to cube both sides:

[(7x + 1)^1/3]^3 = (1+ x)^3

We'll remove the brackets:

7x + 1 = 1 + x^3 + 3x(x+1)

We'll remove the brackets form the right side:

7x + 1 = 1 + x^3 + 3x^2 + 3x

We'll subtract 7x + 1 both sides and we'll use symmetric property:

x^3 + 3x^2 + 3x - 7x + 1 - 1 = 0

We'll eliminate and combine like terms:

x^3 + 3x^2 - 4x = 0

We'll factorize by x:

x(x^2 + 3x - 4) = 0

We'll set each factor as zero:

x = 0

x^2 + 3x - 4 = 0

x^2 + 3x - 3 - 1 = 0

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

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

We'll factorize by x-1:

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

x - 1 = 0

x = 1

x + 4 = 0

x = -4

**The 3 roots of the equation are: {-4 ; 0 ; 1}.**