Question 779410
{{{(2x)/(x^3-3x^2)}}} * {{{(x^2-x-6)/n}}} = {{{1/(3x)}}}
Factor
{{{(2x)/(x^2(x-3))}}} * {{{((x-3)(x+2))/n}}} = {{{1/(3x)}}}
Cancel x and (x-3)
{{{(2)/(x)}}} * {{{((x+2))/n}}} = {{{1/(3x)}}} 
which is:
{{{(2(x+2))/(nx)}}}  = {{{1/(3x)}}}
cross multiply
nx = 2(x+2) * 3x
nx = (2x + 4) * 3x
nx = 6x^2 + 12x
divide both sides by x
n = (6x + 12)
therefore our equation is:
{{{(2x)/(x^3-3x^2)}}} * {{{(x^2-x-6)/(6x+12)}}} = {{{1/(3x)}}}
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You can check this out, factor the above
{{{(2x)/(x^2(x-3))}}} * {{{((x-3)(x+2))/(6(x+2))}}} = {{{1/(3x)}}}
Cancel x, (x-3), and (x+2)
{{{(2)/(x)}}} * {{{1/6}}} = {{{1/(3x)}}} 
which is
{{{2/(6x)}}} = {{{1/(3x)}}}; confirms our answer of n = (6x+12)