# How do you simplify \frac { x + 2} { ( x - 3) ( x + 1) } + \frac { x ( x - 2) } { 3( x - 3) } + \frac { 1- x ^ { 3} } { 3( x ^ { 2} - 2x - 3) } = - \frac { x } { 3( x + 1) }?

Jul 7, 2018

$x = \frac{7}{2}$

#### Explanation:

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

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

$\frac{3 x + 6 + {x}^{3} - {x}^{2} - 2 x + 1 - {x}^{3}}{3 \cdot \left({x}^{2} - 2 x - 3\right)} = - \frac{x}{3 \cdot \left(x + 1\right)}$

$\frac{- {x}^{2} + x + 7}{3 \cdot \left({x}^{2} - 2 x - 3\right)} = - \frac{x}{3 \cdot \left(x + 1\right)}$

$\frac{- {x}^{2} + x + 7}{{x}^{2} - 2 x - 3} = - \frac{x}{x + 1}$

$- \frac{{x}^{2} - x - 7}{\left(x + 1\right) \cdot \left(x - 3\right)} = - \frac{x}{x + 1}$

$\frac{{x}^{2} - x - 7}{\left(x + 1\right) \cdot \left(x - 3\right)} = \frac{x}{x + 1}$

${x}^{2} - x - 7 = x \cdot \left(x - 3\right)$

${x}^{2} - x - 7 = {x}^{2} - 3 x$

$2 x = 7$, so $x = \frac{7}{2}$