Why computation of $u*v^3*(u*v^7)^{(p-5)/8}$ is suggested instead of $(u/v)^{(p+3)/8}$ Working with Curve25519 I've faced with suggested form of computation square root candidate as: $uv^3(uv^7)^{\frac{p-5}{8}}$ instead of $\left(\frac{u}{v}\right)^{\frac{p+3}{8}}$. Why it is so? Or why is it less expensive?

Edwards-Curve Digital Signature Algorithm (EdDSA) https://tools.ietf.org/html/rfc8032#section-5.1.3
High-speed high-security signatures https://eprint.iacr.org/2011/368.pdf
Note that
$$\frac{7(p-5)}8+3=\frac{7p-11}8\equiv \frac{7p-11}8-p-1\equiv -\frac{(p+3)}8\pmod{p-1}.$$
Similarly
$$\frac{p-5}8+1=\frac{p+3}8$$
And so $v^3(v^7)^{\frac{p-5}8}\equiv v^{-\frac{(p+3)}8}\pmod p$.
Likewise $u(u)^{\frac{p-5}8}\equiv u^{\frac{p+3}8}$ as required.
The reason is to save a modular division which is quite a pricey operation.