3. If $c$ and $d$ are two non-zero planar vectors that are not collinear, then among the four pairs of $\boldsymbol{a}, \boldsymbol{b}$ given below, the pair that is not collinear is $(\quad$.
(A) $a=-2(c+d), b=2(c+d)$
(B) $a=c-d, b=-2 c+2 d$
(C) $a=4 c-\frac{2}{5} d, b=c-\frac{1}{10} d$
(D) $a=c+d, b=2 c-2 d$
3. D.

From $a=-2(c+d)=-b$, we know that $a$ and $b$ are collinear, thus eliminating option (A);
$$
a=c-d=-\frac{1}{2}(-2 c+2 d)=-\frac{1}{2} b \text{, so } a \text{ and } b
$$

are collinear, eliminating option (B);
$$
a=4 c-\frac{2}{5} d=4\left(c-\frac{1}{10} d\right)=4 b \text{, so } a \text{ and } b \text{ are collinear, }
$$

eliminating option (C).
In fact, $c$ and $d$ are two non-collinear planar vectors. Then $\boldsymbol{a}=$ $c+d$ is a non-zero vector. If $a$ and $b$ are collinear, there exists a unique real number $\lambda$ such that $\boldsymbol{a}=\lambda \boldsymbol{b}$, hence
$$
c+d=\lambda(2 c-2 d) \Leftrightarrow(1-2 \lambda) c=-(2 \lambda+1) d .
$$

When $1-2 \lambda=0$, $d$ is a zero vector, which contradicts the given information. When $1-2 \lambda \neq 0$, $c=\frac{2 \lambda+1}{2 \lambda-1} d$, indicating that $c$ and $d$ are collinear non-zero planar vectors, which still contradicts the given information. Therefore, $\boldsymbol{a}$ and $\boldsymbol{b}$ are not collinear.