Question
$\frac{1}{{\sqrt[3]{{6 - 3x}}}} = $
$ = {6^{ - 1/3}}\left[ {1 + \left( { - \frac{1}{3}} \right)\,\left( { - \frac{x}{2}} \right)x + \frac{{\left( { - \frac{1}{3}} \right)\,\left( { - \frac{4}{3}} \right)}}{{2.1}}{{\left( { - \frac{x}{2}} \right)}^2} + ....} \right]$
$ = {6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^{^2}}}} + ....} \right]$
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