Divide (x^3+1)÷(x^2-x+1)

$(x^{3} + 1) \div (x^{2} - x + 1)$

Step 1

Rewrite the division as a fraction.

$\frac{x^{3} + 1}{x^{2} - x + 1}$

Step 2

Simplify the numerator.

Rewrite $1$ as $1^{3}$ .

$\frac{x^{3} + 1^{3}}{x^{2} - x + 1}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

$\frac{(x + 1) (x^{2} - x \cdot 1 + 1^{2})}{x^{2} - x + 1}$

Simplify.

Multiply $- 1$ by $1$ .

$\frac{(x + 1) (x^{2} - x + 1^{2})}{x^{2} - x + 1}$

One to any power is one.

$\frac{(x + 1) (x^{2} - x + 1)}{x^{2} - x + 1}$

Step 3

Cancel the common factor of $x^{2} - x + 1$ .

Cancel the common factor.

$\frac{(x + 1) (x^{2} - x + 1)}{x^{2} - x + 1}$

Divide $x + 1$ by $1$ .

$x + 1$

Divide (x^3+1)÷(x^2-x+1)

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