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

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

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

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

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

x = 0, x = 1

Solution steps

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

Expand

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

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

Move

1

to the left side

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

Move

x^{3}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 0 = 2

x_{1, 2} = \frac{- 2 \pm 2}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 2 + 2}{2 ( - 2 )}, x_{2} = \frac{- 2 - 2}{2 ( - 2 )}

x = \frac{- 2 + 2}{2 ( - 2 )} : 0

x = \frac{- 2 - 2}{2 ( - 2 )} : 1

The solutions to the quadratic equation are:

x = 0, x = 1

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

x, \frac{( x ^{2} + m ^{2} )}{v} = \frac{( v ^{2} - s )}{m}

x, x^{2} + y^{2} + z^{2} = x + y + z + 84

x, [3 p + 1] \cdot x^{2} + 2 [p + 1] \cdot x + p = 0

\frac{m}{n} \cdot x^{2} + \frac{n}{m} = 1 - 2 x

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