Solution
[tex]\frac{1}{3x^2}-\frac{x}{x^2-8}[/tex]Find the LCM and subtarct the numerator
[tex]\text{LCM }=3x^2(x^2-8)\text{ }[/tex][tex]\begin{gathered} =\frac{\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)}{3x^2\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)}-\frac{3x^3}{3x^2\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)} \\ \mathrm{Apply\: the\: fraction\: rule}\colon\quad \frac{a}{c}-\frac{b}{c}=\frac{a-b}{c} \end{gathered}[/tex][tex]\begin{gathered} =\frac{\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)-3x^3}{3x^2\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)} \\ =\frac{x^2-8-3x^3}{3x^2\left(x+2\sqrt{2}\right)\left(x-2\sqrt{2}\right)} \end{gathered}[/tex][tex]\begin{gathered} Expand(x+2\sqrt[]{2)}(x-2\sqrt[]{2)} \\ =\frac{x^2-8-3x^3}{3x^2\left(x^2-8\right)} \end{gathered}[/tex]
