Divide (x+x^4)/(x^2-x+1)

$\frac{x + x^{4}}{x^{2} - x + 1}$

Simplify the numerator.

Factor $x$ out of $x + x^{4}$ .

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

$\frac{x^{1} + x^{4}}{x^{2} - x + 1}$

Factor $x$ out of $x^{1}$ .

$\frac{x \cdot 1 + x^{4}}{x^{2} - x + 1}$

Factor $x$ out of $x^{4}$ .

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

Factor $x$ out of $x \cdot 1 + x \cdot x^{3}$ .

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

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

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

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 $a = 1$ and $b = x$ .

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

Simplify.

One to any power is one.

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

Rewrite $- 1 x$ as $- x$ .

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x (1 + x)$ by $1$ .

$x (1 + x)$

Apply the distributive property.

$x \cdot 1 + x \cdot x$

Multiply $x$ by $1$ .

$x + x \cdot x$

Multiply $x$ by $x$ .

$x + x^{2}$

Divide (x+x^4)/(x^2-x+1)

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