Let $a$ be a natural number such that $\gcd (a, 16) = 1$.
Find the multiplicative inverse of $a^3$ modulo 16 as a power of $a$.
Since $\gcd(a,16)=1$, Euler's Theorem gives us
$$ a^{\phi(16)} \equiv 1 \pmod {16} $$
Using $\phi(16)=8$,
$$ a^{8} \equiv 1 \pmod {16} $$
Factorising,
$$ a^3 \times a^5 \equiv 1 \pmod {16} $$
And so the multiplicative inverse of $a^3$ is
$$ (a^3)^{-1} \equiv a^5 \pmod {16} $$