Q. The expression $\frac{1}{\sqrt[3]{6 - 3x}}$ is equal to
A) $6^{\frac{1}{3}}\left\lbrack 1 + \frac{x}{6} + \frac{{2x}^{2}}{6^{2}} + ... \right\rbrack$
B) $6^{- \frac{1}{3}}\left\lbrack 1 + \frac{x}{6} + \frac{{2x}^{2}}{6^{2}} + ... \right\rbrack$
C) $6^{\frac{1}{3}}\left\lbrack 1 - \frac{x}{6} + \frac{{2x}^{2}}{6^{2}} - ... \right\rbrack$
D) $6^{- \frac{1}{3}}\left\lbrack 1 - \frac{x}{6} + \frac{{2x}^{2}}{6^{2}} - ... \right\rbrack$
Explanation
(b)
$\frac{1}{{(6 - 3x)}^{\frac{1}{3}}} = {(6 - 3x)}^{\frac{- 1}{3}} = 6^{- \frac{1}{3}}\left\lbrack 1 - \frac{x}{2} \right\rbrack^{\frac{- 1}{3}}$
$= 6^{\frac{- 1}{3}}\left\lbrack 1 + \left( - \frac{1}{3} \right)\left( - \frac{x}{2} \right) + \frac{\left( - \frac{1}{3} \right)\left( - \frac{4}{3} \right)}{2.1}\left( - \frac{x}{2} \right)^{2} + ... \right\rbrack$
$= 6^{\frac{- 1}{3}}\left\lbrack 1 + \frac{x}{6} + \frac{{2x}^{2}}{6^{2}} + ... \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.