# xFind x if (7x+1)^1/3-x=1 .

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You need to isolate the irrational term to one side, such that:

`root(3)(7x + 1) = x + 1`

You need to remove the cube root, hence, you need to raise the equation to 3rd power, such that:

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

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

Reducing duplicate terms yields:

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

Factoring out x to the right yields:

`7x = x(x^2 + 3x + 3)`

`7x - x(x^2 + 3x + 3) = 0 => x(7 - x^2 - 3x - 3) = 0`

`x(x^2 + 3x - 4) = 0 => {(x = 0),(x^2 + 3x - 4 = 0):}`

You need to use quadratic formula to solve for x the equation `x^2 + 3x - 4 = 0` , such that:

`x^2 + 3x - 4 = 0 => x_(1,2) = (-3+-sqrt(3^2 - 4*1*(-4)))/(2*1)`

`x_(1,2) = (-3 +- 5)/2 => x_1 = 1 ; x_2 = -4`

**Hence, evaluating the solutions to the given equation yields **`x_1=0,x_2=1,x_3=-4.`

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}.