$$\frac { 1 } { x ^ { 2 } + 6 x + 8 } + \frac { 2 } { x ^ { 2 } - 4 } = \frac { 3 } { x ^ { 2 } + 2 x - 8 }$$
Solve for x (complex solution)
$x\in \mathrm{C}\setminus -4,-2,2$
Solution Steps
Variable $x$ cannot be equal to any of the values $-4,-2,2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-2\right)\left(x+2\right)\left(x+4\right)$, the least common multiple of $x^{2}+6x+8,x^{2}-4,x^{2}+2x-8$.
Variable $x$ cannot be equal to any of the values $-4,-2,2$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-2\right)\left(x+2\right)\left(x+4\right)$, the least common multiple of $x^{2}+6x+8,x^{2}-4,x^{2}+2x-8$.
