Solve for x x^3+1=x^2+x

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

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Subtract $x^{3}$ from both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

Factor the left side of the equation.

Reorder terms.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

Combine exponents.

Factor $- 1$ out of $- x$ .

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

Rewrite $1$ as $- 1 (- 1)$ .

$(- (x) - 1 \cdot - 1) (x + 1) (x - 1) = 0$

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

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

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

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

Raise $x - 1$ to the power of $1$ .

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

Raise $x - 1$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 1 {(x - 1)}^{1 + 1} (x + 1) = 0$

Add $1$ and $1$ .

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 1)}^{2} = 0$

$x + 1 = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

${(x - 1)}^{2} = 0$

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- 1 {(x - 1)}^{2} (x + 1) = 0$ true.

$x = 1, - 1$

Solve for x x^3+1=x^2+x

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