TeX source:
\begin{array}{rclcl}0 &=& \dfrac{\sqrt[3]{\left(x^2+8x+16\right)^3}}{\sqrt{17+x}} \\\\0 &=& \dfrac{x^2+8x+16}{\sqrt{17+x}} \\\\0 &=& \dfrac{\left(x+4\right)^2}{\sqrt{17+x}} &\vert& \cdot \sqrt{17+x} \\\\0 &=& \left(x+4\right)^2 &\vert& \pm\sqrt{} \\ 0 &=& x+4 &\vert& -4 \\ x &=& -4 \in \mathbb{D} \end{array}