3. Given $a, b, c \in \mathbf{R}_{+}$, and $abc=4$. Then the minimum value of the algebraic expression $a^{a+b} b^{3b} c^{c+b}$ is $\qquad$

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3. 64.

It is easy to know that $a^{a+b} b^{3 b} c^{c+b}=a^{a} b^{2 b} c^{c} \cdot 4^{b}=a^{a}(2 b)^{2 b} c^{c}$.
And $a b c=4$ can be transformed into $a \cdot 2 b \cdot c=8$, considering $2 b$ as a whole.

Since the function $f(x)=\ln x$ is increasing on $(0,+\infty)$, for any $a, b \in(0,+\infty)$, we always have $(a-b)(f(a)-f(b)) \geqslant 0$, i.e., $a \ln a+b \ln b \geqslant a \ln b+b \ln a$.
Substituting $2 b$ for $b$ in the above inequality, we get $a \ln a+2 b \ln 2 b \geqslant a \ln 2 b+2 b \ln a$.
Similarly, $2 b \ln 2 b+c \ln c \geqslant 2 b \ln c+c \ln 2 b$, (2) $c \ln c+a \ln a \geqslant c \ln a+a \ln c$.
(1) + (2) + (3) and rearranging gives
$3 \ln \left[a^{a}(2 b)^{2 b} c^{c}\right] \geqslant(a+2 b+c) \ln (a \cdot 2 b \cdot c)$,

i.e., $a^{a}(2 b)^{2 b} c^{c} \geqslant(a \cdot 2 b \cdot c)^{\frac{a+2 b+c}{3}} \geqslant 8 \sqrt[3 / 2 \cdot 2 b \cdot c]{ }=64$.
The equality holds if and only if $a=c=2, b=1$. Therefore, the minimum value of $a^{a+b} b^{3 b} c^{c+b}$ is 64.