Answer
$1+3x^2+3x^4+x^6$
Work Step by Step
The formula to determine the binomial series is:
$(1+x)^p=1+\Sigma_{k=1}^\infty \dbinom{p}{k}x^k$
Here, $\dbinom{p}{k}=\dfrac{p(p-1)(p-2).....(p-k+1)}{k!}$
Now, $(1+x^2)^{3}=1+3x^2+\dfrac{(3)(2)(x^2)^2}{2!}+\dfrac{(3)(2)(1)(x^2)^3}{3!}+0=1+3x^2+3x^4+x^6$
