How would I prove or disprove this?
I tried it like this:
$$8^{n-3}\in O\left(2^n\right)$$ $$8^{n-3}\leq c\cdot 2^n$$ $$8^n\cdot8^{-3}\leq c\cdot 2^n$$ $$8^n\leq c\cdot 2^n$$
(We can remove the $8^{-3}$ because it is a constant)
But I'm really unsure. Is this even correct? What would c have to be then?
Hint:
$8^n = 2^n\cdot 4^n$. Think about the $4^n$ and how it can outgrow any constant $C$.
$8^{n-3}=O(2^n)$ means that the sequence $a_n= \frac{8^{n-3}}{2^n}$ is bounded. But $a_n \to \infty$ as $n \to \infty.$
Consequence: $8^{n-3}=O(2^n)$ is not true.
Divide both sides by $2^n$ and you should get $$\frac{8^n}{2^n}=\left(\frac82\right)^n=4^n\leq c$$ Since $\lim_{n\to\infty}4^n=\infty$, i.e. $4^n$ diverges, there cannot exist constant $c$ such that $4^n\leq c$ for all $n$.
Therefore $8^{n-3}\notin O\left(2^n\right)$
It is false that $8^{n-3} = O(2^n)$. Indeed, $$\limsup_{n\to\infty}\frac{8^{n-3}}{2^n} = \limsup_{n\to\infty} 2^{2n-9} = \infty,$$ which contradicts the definition of $f=O(g)$, i.e., that $$\limsup_{n\to\infty}\frac{|f(n)|}{g(n)}<\infty.$$
Taking the base-$2$ logarithm,
$$3(n-3)\le\log_2c+n$$ or $$\log_2c\ge 2n-6$$ is certainly not possible.