# How do you simplify (x^3 - 64)/(x^3 + 64) div (x^2 - 16)/(x^2 - 4x + 16)?

Jul 14, 2015

$\frac{{x}^{3} - 64}{{x}^{3} + 64} \div \frac{{x}^{2} - 16}{{x}^{2} - 4 x + 16}$

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

#### Explanation:

$\frac{{x}^{3} - 64}{{x}^{3} + 64} \div \frac{{x}^{2} - 16}{{x}^{2} - 4 x + 16}$

$= \frac{{x}^{3} - {4}^{3}}{{x}^{3} + {4}^{3}} \cdot \frac{{x}^{2} - 4 x + {4}^{2}}{{x}^{2} - {4}^{2}}$

$= \frac{\cancel{x - 4} \left({x}^{2} + 4 x + {4}^{2}\right)}{\left(x + 4\right) \cancel{{x}^{2} - 4 x + {4}^{2}}} \cdot \frac{\cancel{{x}^{2} - 4 x + {4}^{2}}}{\cancel{x - 4} \left(x + 4\right)}$

$= \frac{{x}^{2} + 4 x + {4}^{2}}{x + 4} ^ 2$

$= \frac{\left({x}^{2} + 8 x + {4}^{2}\right) - 4 x}{x + 4} ^ 2$

$= \frac{{\left(x + 4\right)}^{2} - 4 x}{x + 4} ^ 2$

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

with exclusion $x \ne 4$ (coming from the cancelled $\left(x - 4\right)$ term)

..using difference of cubes:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

..sum of cubes:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

..and difference of squares:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$