According to Wikipedia,
The degree of a differential equation is the power of its highest derivative, after the equation has been made rational and integral in all of its derivatives.
Using this definition we find that the degree of the following ODE is 2. $$\left(\frac {d^3y}{dx^3}\right)^2-\frac {dy}{dx}+5=0$$ We can also express the above ODE as follows, $$\left(\frac {d^3y}{dx^3}\right)^6-\left(\frac {dy}{dx}-5\right)^3=0$$ But using the same definition we get that the degree is 6. Can someone explain why the degree is not 6.
Edit-1
Instead of squaring I cubed the initial ODE.
Edit-2
I understood from the comments that they are different DE, so is there any point in asking to find the degree of $$\left(\frac {d^3y}{dx^3}\right)-\sqrt {\frac {dy}{dx}+5}=0$$
A DE has the form of an algebraic equation of degree $m$ in the highest derivative, then we say that the given DE is of degree $m$. Regarding your question in $\textbf{Edit-2}.$ Rewrite the equation $\displaystyle \left(\frac{d^3y}{dx^3}\right)=\sqrt{\frac{dy}{dx}-5}$. Now taking $2m\, (m \in \mathbb{N})$ powers both sides we obtain $\displaystyle \left(\frac{d^3y}{dx^3}\right)^{2m}=\left(\frac{dy}{dx}-5\right)^m$. This a polynomial in $\displaystyle \frac{d^3y}{dx^3}$ with highest degree $2m$ for each $m\in \mathbb{N}$. So degree of the above differential equation is $2m$. For each $m\in \mathbb{N}$, $2m$ is a degree. But, degree of a polynomial is unique. To fix the uniqueness of the degree, take the lowest power which reduces the equation into a polynomial. Thus, the degree of the above DE is 2.