# Fiat Shamir with S^3

I'm trying to do Fiat Shamir but with $I = S^3 \pmod n$ where $S$ is the secret key. How will I prove that I know the cube root? I tried doing the old formula but with instead with $(RS)^3 = x*I \pmod n$ where $R$ is the random number and $x = R^2 \pmod n$, but it didn't seem to yield the correct values. Só what formula should I be using instead with the new cube.

$x = R^2 \pmod n$

$R^2 = x \pmod n$

$(RS)^2 = x*I \pmod n$

I tried changing these formulas to use cubes instead but didn't work.