What are the solutions to the equation (x(x+1)(x-1))3=2x-2 ?

The solutions to the equation (x(x+1)(x-1))3=2x-2 are x=1,x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6

Find the zeros of (x(x+1)(x-1))3=2x-2

The zeros of (x(x+1)(x-1))3=2x-2 are x=1,x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6

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

x = 1, x = \frac{- 3 + 33}{6}, x = \frac{- 3 - 33}{6}

Decimal Notation

x = 1, x = 0.45742 \dots, x = - 1.45742 \dots

Solution steps

(x (x + 1) (x - 1)) \cdot 3 = 2 x - 2

Expand

(x (x + 1) (x - 1)) \cdot 3 : 3 x^{3} - 3 x

3 x^{3} - 3 x = 2 x - 2

Move

2

to the left side

3 x^{3} - 3 x + 2 = 2 x

Move

2 x

to the left side

3 x^{3} - 5 x + 2 = 0

Factor

3 x^{3} - 5 x + 2 : (x - 1) (3 x^{2} + 3 x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 3 x^{2} + 3 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x^{2} + 3 x - 2 = 0 : x = \frac{- 3 + 33}{6}, x = \frac{- 3 - 33}{6}

The solutions are

x = 1, x = \frac{- 3 + 33}{6}, x = \frac{- 3 - 33}{6}

2 x^{4} - x^{2} - 15 = 0

3 x^{4} + 4 x^{3} - x^{2} = x^{2}

x^{3} - 17 x^{2} = 0

x^{3} - 12 x^{2} + 48 x - 72 = 0

(4 x^{8}) = 28 x^{10}

What are the solutions to the equation (x(x+1)(x-1))3=2x-2 ?
The solutions to the equation (x(x+1)(x-1))3=2x-2 are x=1,x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6
Find the zeros of (x(x+1)(x-1))3=2x-2
The zeros of (x(x+1)(x-1))3=2x-2 are x=1,x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6