Explanation:
$\frac{2x^4-162}{\left(x^2+9\right)\left(2x-6\right)}$
$=2x^4-162:\quad 2\left(x^4-81\right)$
$=\mathrm{Factor}\:2x-6:\quad 2\left(x-3\right)$
$=\frac{2\left(x^4-81\right)}{2\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1$
$\frac{x^4-81}{\left(x^2+9\right)\left(x-3\right)}$
$\:x^4-81:\quad \left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$x^4-81$
$x^4-9^2$
$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^4-9^2=\left(x^2+9\right)\left(x^2-9\right)$
$=\left(x^2+9\right)\left(x^2-9\right)$
$\mathrm{Factor}\:x^2-9:\quad \left(x+3\right)\left(x-3\right)$
$=\left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$=\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:}\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}:\quad x+3$
$\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:the\:common\:factor:}\:x-3$
$\frac{\left(x+3\right)\left(x^2+9\right)}{x^2+9}$
$\mathrm{Cancel\:the\:common\:factor:}\:x^2+9$
$=x+3$