# How do you simplify (x^2-16)/x^3+(x^3+1)/x^4?

Aug 19, 2017

See a solution process below:

#### Explanation:

To add fractions they must be over a common denominator. We can multiply the fraction on the left by the appropriate form of $1$ to give it a common denominator with the fraction on the right:

$\left(\frac{x}{x} \times \frac{{x}^{2} - 16}{x} ^ 3\right) + \frac{{x}^{3} + 1}{x} ^ 4 \implies$

$\frac{x \left({x}^{2} - 16\right)}{x \times {x}^{3}} + \frac{{x}^{3} + 1}{x} ^ 4 \implies$

$\frac{\left(x \times {x}^{2}\right) - \left(x \times 16\right)}{x} ^ 4 + \frac{{x}^{3} + 1}{x} ^ 4 \implies$

$\frac{{x}^{3} - 16 x}{x} ^ 4 + \frac{{x}^{3} + 1}{x} ^ 4$

We can now add the numerators of the two fractions over the common denominator:

$\frac{{x}^{3} - 16 x + {x}^{3} + 1}{x} ^ 4 \implies$

$\frac{{x}^{3} + {x}^{3} - 16 x + 1}{x} ^ 4 \implies$

$\frac{1 {x}^{3} + 1 {x}^{3} - 16 x + 1}{x} ^ 4 \implies$

$\frac{\left(1 + 1\right) {x}^{3} - 16 x + 1}{x} ^ 4 \implies$

$\frac{2 {x}^{3} - 16 x + 1}{x} ^ 4$