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

Apr 4, 2015

$\frac{4}{3 {\left(x + 4\right)}^{3}} \cdot \left[\frac{1}{x + 2} ^ 2 - \frac{x + 2}{x + 4} ^ 4\right]$

Given:
$\frac{\frac{{\left(x + 4\right)}^{3}}{4} \cdot {\left(x + 2\right)}^{-} \frac{2}{3} - \frac{x + 2}{3} \cdot {\left(x + 4\right)}^{-} \frac{1}{4}}{{\left(x + 4\right)}^{3} / 4} ^ 2$

Let $a = x + 4$ and $b = x + 2$
$\frac{\frac{{a}^{3}}{4} \cdot {b}^{-} \frac{2}{3} - \frac{b}{3} \cdot {a}^{-} \frac{1}{4}}{{a}^{3} / 4} ^ 2$

When you divide by a fraction, you are multiplying the reciprocal:
$\left[\frac{{a}^{3}}{4} \cdot {b}^{-} \frac{2}{3} - \frac{b}{3} \cdot {a}^{-} \frac{1}{4}\right] \cdot {\left(\frac{4}{a} ^ 3\right)}^{2}$

Change negative exponents to reciprocals with positive exponents:
$\left[\frac{{a}^{3}}{4} \cdot \frac{1}{3 {b}^{2}} - \frac{b}{3} \cdot \frac{1}{4 a}\right] \cdot {\left(\frac{4}{a} ^ 3\right)}^{2}$

$\left[{a}^{3} / \left(12 {b}^{2}\right) - \frac{b}{12 a}\right] \left(\frac{16}{a} ^ 6\right)$

$\frac{16 {a}^{3}}{12 {a}^{6} {b}^{2}} - \frac{16 b}{12 {a}^{7}}$

$\frac{4}{3 {a}^{3} {b}^{2}} - \frac{4 b}{3 {a}^{7}}$

Factor out $\frac{4}{3 {a}^{3}}$:

$\frac{4}{3 {a}^{3}} \left[\frac{1}{b} ^ 2 - \frac{b}{a} ^ 4\right]$

Substitute back in $x + 4$ and $x + 2$
$\frac{4}{3 {\left(x + 4\right)}^{3}} \cdot \left[\frac{1}{x + 2} ^ 2 - \frac{x + 2}{x + 4} ^ 4\right]$