The equation of the line normal to the function $f(x) = \left(x – 8\right)^{2/3} + 1$ at $P(0, 5)$ is $3y = x – 15$.
The slope of the tangent line to the function $f(x)$ at $x = a$ is given by $f'(a)$. The slope of the normal line to the function $f(x)$ at $x = a$ is $-\frac{1}{f'(a)}$.
In this case, $f'(x) = \frac{2}{3} \left(x – 8\right)^{-1/3}$. Therefore, the slope of the tangent line to the function $f(x)$ at $x = 0$ is $2$.
The normal line to the function $f(x)$ at $x = 0$ has a slope of $-\frac{1}{2}$. The equation of the line with slope $-\frac{1}{2}$ and passing through the point $(0, 5)$ is $y – 5 = -\frac{1}{2}(x – 0)$, which simplifies to $3y = x – 15$.
Therefore, the equation of the line normal to the function $f(x) = \left(x – 8\right)^{2/3} + 1$ at $P(0, 5)$ is $3y = x – 15$.