Simplify (16a^4+16a)/(a^2+1-a)

$\frac{16 a^{4} + 16 a}{a^{2} + 1 - a}$

Simplify the numerator.

Factor $16 a$ out of $16 a^{4} + 16 a$ .

Factor $16 a$ out of $16 a^{4}$ .

$\frac{16 a (a^{3}) + 16 a}{a^{2} + 1 - a}$

Factor $16 a$ out of $16 a$ .

$\frac{16 a (a^{3}) + 16 a (1)}{a^{2} + 1 - a}$

Factor $16 a$ out of $16 a (a^{3}) + 16 a (1)$ .

$\frac{16 a (a^{3} + 1)}{a^{2} + 1 - a}$

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

$\frac{16 a (a^{3} + 1^{3})}{a^{2} + 1 - a}$

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

$\frac{16 a ((a + 1) (a^{2} - a \cdot 1 + 1^{2}))}{a^{2} + 1 - a}$

Simplify.

Multiply $- 1$ by $1$ .

$\frac{16 a ((a + 1) (a^{2} - a + 1^{2}))}{a^{2} + 1 - a}$

One to any power is one.

$\frac{16 a ((a + 1) (a^{2} - a + 1))}{a^{2} + 1 - a}$

$\frac{16 a (a + 1) (a^{2} - a + 1)}{a^{2} + 1 - a}$

Simplify terms.

Reorder terms.

$\frac{16 a (a + 1) (a^{2} - a + 1)}{a^{2} - a + 1}$

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

Cancel the common factor.

$\frac{16 a (a + 1) (a^{2} - a + 1)}{a^{2} - a + 1}$

Divide $16 a (a + 1)$ by $1$ .

$16 a (a + 1)$

Apply the distributive property.

$16 a \cdot a + 16 a \cdot 1$

Multiply $a$ by $a$ by adding the exponents.

Move $a$ .

$16 (a \cdot a) + 16 a \cdot 1$

Multiply $a$ by $a$ .

$16 a^{2} + 16 a \cdot 1$

Multiply $16$ by $1$ .

$16 a^{2} + 16 a$

Simplify (16a^4+16a)/(a^2+1-a)

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