# How do you use the binomial series to expand 1/sqrt(1-x^3)?

Jul 3, 2016

${\left(1 - {x}^{3}\right)}^{- \frac{1}{2}} = 1 + \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} + \frac{5}{16} {x}^{9} + \frac{35}{128} {x}^{12} + \ldots \ldots .$

#### Explanation:

$\frac{1}{\sqrt{1 - {x}^{3}}} = {\left(1 - {x}^{3}\right)}^{- \frac{1}{2}}$

According to binomial series (1-a)^n=1-na+(n(n-1))/(2!)a^2-(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4-.......

Hence (1-x^3)^(-1/2)=1-(-1/2)x^3+((-1/2)(-1/2-1))/(2!)x^6-((-1/2)(-1/2-1)(-1/2-2))/(3!)x^9+((-1/2)(-1/2-1)(-1/2-2)(-1/2-3))/(4!)x^12-.......

or (1-x^3)^(-1/2)=1+1/2x^3+((-1/2)(-3/2))/(2!)x^6-((-1/2)(-3/2)(-5/2))/(3!)x^9+((-1/2)(-3/2)(-5/2)(-7/2))/(4!)x^12+.......

or ${\left(1 - {x}^{3}\right)}^{- \frac{1}{2}} = 1 + \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} + \frac{15}{48} {x}^{9} + \frac{105}{384} {x}^{12} + \ldots \ldots .$

or ${\left(1 - {x}^{3}\right)}^{- \frac{1}{2}} = 1 + \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} + \frac{5}{16} {x}^{9} + \frac{35}{128} {x}^{12} + \ldots \ldots .$