Here we will use the standard binomial series Representation for the factorial power in order to determine the power series representation for the given function.First of all we convert the given function into the standard form then we will go for series representation an the interval of convergence.

\((1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)

The function for which we have to determine the power series representation is given as

\(f(x)=\sqrt{1+x^3}\)

Let us assume \(a=x^3\),

now

\(\sqrt{1+a}=(1+a)^{\frac{1}{2}}=1+\frac{a}{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}a^2+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}a^3+...\)

the above series is valid only when \(|a|<1\)</span>

\(\Rightarrow(1+a)^{\frac{1}{2}}=1+\frac{a}{2}-\frac{a^2}{8}+\frac{a^3}{16}+...\)

The power series representation is given as

\(f(x)=\sqrt{1+x^3}=1+\frac{x^3}{2}-\frac{x^6}{8}+\frac{x^9}{16}+...\)

The interval of convergence is given as \(|x^3|<1\Rightarrow|x|<1\Rightarrow R=1\)</span>

\((1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)

The function for which we have to determine the power series representation is given as

\(f(x)=\sqrt{1+x^3}\)

Let us assume \(a=x^3\),

now

\(\sqrt{1+a}=(1+a)^{\frac{1}{2}}=1+\frac{a}{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}a^2+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}a^3+...\)

the above series is valid only when \(|a|<1\)</span>

\(\Rightarrow(1+a)^{\frac{1}{2}}=1+\frac{a}{2}-\frac{a^2}{8}+\frac{a^3}{16}+...\)

The power series representation is given as

\(f(x)=\sqrt{1+x^3}=1+\frac{x^3}{2}-\frac{x^6}{8}+\frac{x^9}{16}+...\)

The interval of convergence is given as \(|x^3|<1\Rightarrow|x|<1\Rightarrow R=1\)</span>