If $\mathrm{f}$ is one-to-one, find an equation for its inverse. \[ f(x)=x^{3}-6 \] A. $f^{-1}(x)=\sqrt[3]{x-6}$ B. $f^{-1}(x)=\sqrt[3]{x}+6$ C. $f^{-1}(x)=\sqrt[3]{x+6}$ D. The function is not one-to-one.

Step 1 ：Given that the function \(f(x)=x^{3}-6\) is one-to-one, we can find its inverse by swapping \(x\) and \(y\) and solving for \(y\).

Step 2 ：So, we have \(x=y^{3}-6\).

Step 3 ：Adding 6 to both sides, we get \(x+6=y^{3}\).

Step 4 ：Taking the cube root of both sides, we get \(y=\sqrt[3]{x+6}\).

Step 5 ：So, the inverse function \(f^{-1}(x)=\sqrt[3]{x+6}\).

Step 6 ：Comparing this with the options given, we find that the correct answer is \(\boxed{\text{C. } f^{-1}(x)=\sqrt[3]{x+6}}\).