Answer
$x^k-3$
Work Step by Step
The given expression can be written as:
$\\=\dfrac{(x^k)^2-3^2}{3+x^k}
\\=\dfrac{(x^k)^2-3^2}{x^k+3}$
RECALL:
$a^2-b^2=(a-b)(a+b)$
Factor the numerator using the formula above where $a=x^k$ and $b=3$ to have:
$\\=\dfrac{(x^k-3)(x^k+3)}{x^k+3}$
Cancel the common factors to have:
$\\\require{cancel}=\dfrac{(x^k-3)\cancel{(x^k+3)}}{\cancel{x^k+3}}
\\=x^k-3$
