One Theorem and example in Recursive Formula I read a book as follows:

and following example:
$T(n)=T(\frac{n}{4})+O(\log^2n)=O(\log^3n)$
How we can fit this theorem in this example? Or this is cannot be handle in this way what is the most good method for solving this reuccrence?
Update: I need more quick version for finding solution of this recurrence.
 A: Assuming feasible $a,b,c$, a simple way to solve this recurrence:
$$
T(n) =a T\left(\frac nb\right)+n^c
$$
$$
T\left(b^{\log_b n}\right)= aT\left(b^{\log_b\frac nb}\right)+n^c
$$
now making
$$
\mathcal{T}(\cdot)=T\left(b^{(\cdot)}\right),\ \ \ z=\log_b n
$$
$$
\mathcal{T}(z)= a \mathcal{T}(z-1)+b^{zc}
$$
which is a linear recurrence with solution
$$
\mathcal{T}(z) = \mathcal{T}(z)_h + \mathcal{T}(z)_p,\ \ \cases{\mathcal{T}(z)_h=a \mathcal{T}(z-1)_h\\
\mathcal{T}(z)_p=a \mathcal{T}(z-1)_p+b^{zc}}
$$
The homogeneous has as solution
$$
\mathcal{T}(z-1)_h=a^{z-1}C_0
$$
for the particular, making $\mathcal{T}(z-1)_p=a^{z-1}C_0(z)$ and substituting we have
$$
C_0(z)-C_0(z-1) = \frac{b^{z c}}{a^{z-1}}
$$
and then
$$
C_0(z) = \frac{a^{1-z} \left(b^{c (z+1)}-a^z b^c\right)}{b^c-a}
$$
so
$$
\mathcal{T}(z)_p = \frac{b^{c (z+1)}-a^z b^c}{b^c-a}
$$
hence
$$
\mathcal{T}(z) = C_0 a^{z-1}+\frac{b^{c (z+1)}-a^z b^c}{b^c-a}
$$
now making the backwards substitution  $z\to\log_b n$ we can recover $T(n)$
$$
T(n) = C_0 a^{\log_b n-1}+\frac{b^c \left(n^c-a^{\log_b n}\right)}{b^c-a}
$$
