1. ## Power Series

A dont understand how the step highlighted in red came about. How did the $\displaystyle 3$ become $\displaystyle 3^2$ and the $\displaystyle 3^2$ become $\displaystyle 3^4$? And, also, how did the $\displaystyle 3^\frac{1}{3}$ before $\displaystyle (1+\frac{y^2}{3})^\frac{1}{3}$ come about when changing the form of $\displaystyle (3+y^2)^\frac{1}{3}$?

2. Originally Posted by Haris
A dont understand how the step highlighted in red came about. How did the $\displaystyle 3$ become $\displaystyle 3^2$ and the $\displaystyle 3^2$ become $\displaystyle 3^4$? And, also, how did the $\displaystyle 3^\frac{1}{3}$ before $\displaystyle (1+\frac{y^2}{3})^\frac{1}{3}$ come about when changing the form of $\displaystyle (3+y^2)^\frac{1}{3}$?
Hi

$\displaystyle \left(3+y^2\right)^\frac{1}{3} = \left(3\:\left(1+\frac{y^2}{3}\right)\right)^\frac {1}{3} = 3^\frac{1}{3} \left(1+\frac{y^2}{3}\right)^\frac{1}{3}$

because $\displaystyle (ab)^n = a^n\:b^n$

Then

$\displaystyle \binom{\frac{1}{3}}{1} = \frac{1}{3}$

Therefore
$\displaystyle \binom{\frac{1}{3}}{1} \frac{1}{3} \:y^2= \frac{y^2}{9}$

Then
$\displaystyle \binom{\frac{1}{3}}{2} = \frac{\frac{1}{3}\:\left(\frac{1}{3}-1\right)}{2} = \frac{\frac{1}{3}\:\left(\frac{-2}{3}\right)}{2} = -\frac{1}{9}$