1 [3] 6 - 3x = - Vidyadip

MCQ

$\frac{1}{{\sqrt[3]{{6 - 3x}}}} = $

A
${6^{1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} + ....} \right]$
✓
${6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} + ....} \right]$
C
${6^{1/3}}\left[ {1 - \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} - ....} \right]$
D
${6^{ - 1/3}}\left[ {1 - \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} - ....} \right]$

✓

Answer

Correct option: B.

${6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} + ....} \right]$

b
(b) $\frac{1}{{{{(6 - 3x)}^{1/3}}}} = {(6 - 3x)^{ - 1/3}} = {6^{ - 1/3}}{\left[ {1 - \frac{x}{2}} \right]^{ - 1/3}}$

$ = {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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