If [tex]f^{-1}(x)[/tex] is the inverse of [tex]f(x)=(x-8)^2[/tex] for [tex]x\ge8[/tex], then for [tex]f^{-1}(x)\ge8[/tex] we have
[tex]f\left(f^{-1}(x)\right) = \left(f^{-1}(x)-8\right)^2 = x \\\\ \sqrt{\left(f^{-1}(x)-8\right)^2} = \sqrt{x} \\\\ f^{-1}(x)-8 = \sqrt{x} \\\\ \boxed{f^{-1}(x) = 8+\sqrt{x}}[/tex]
More precisely,
[tex]\sqrt{\left(f^{-1}(x)-8\right)^2} = \left|f^{-1}(x)-8\right|[/tex]
but since the inverse is at least 8, the quantity inside the absolute is non-negative, so
[tex]\sqrt{\left(f^{-1}(x)-8\right)^2} = f^{-1}(x)-8[/tex]
