What are the solutions to the equation (2x-1)^3+12(x-1)(x+1)=14x-13 ?

The solutions to the equation (2x-1)^3+12(x-1)(x+1)=14x-13 are x=0,x=-1,x=1

Find the zeros of (2x-1)^3+12(x-1)(x+1)=14x-13

The zeros of (2x-1)^3+12(x-1)(x+1)=14x-13 are x=0,x=-1,x=1

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(2 x - 1)^{3} + 12 (x - 1) (x + 1) = 14 x - 13

x = 0, x = - 1, x = 1

Solution steps

(2 x - 1)^{3} + 12 (x - 1) (x + 1) = 14 x - 13

Expand

(2 x - 1)^{3} + 12 (x - 1) (x + 1) : 8 x^{3} + 6 x - 13

8 x^{3} + 6 x - 13 = 14 x - 13

Move

13

to the right side

8 x^{3} + 6 x = 14 x

Move

14 x

to the left side

8 x^{3} - 8 x = 0

Factor

8 x^{3} - 8 x : 8 x (x + 1) (x - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x - 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = 0, x = - 1, x = 1

x^{3} + 6 = - 12

4 x^{3} = 12

x^{4} = 50625

4 x^{3} - 20 x - 9 = 0

3 t - t^{3} = 0

What are the solutions to the equation (2x-1)^3+12(x-1)(x+1)=14x-13 ?
The solutions to the equation (2x-1)^3+12(x-1)(x+1)=14x-13 are x=0,x=-1,x=1
Find the zeros of (2x-1)^3+12(x-1)(x+1)=14x-13
The zeros of (2x-1)^3+12(x-1)(x+1)=14x-13 are x=0,x=-1,x=1