**HINT** $\;$ Work "**generically**", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be **indeterminates** and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the *domain* $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the **universality** of polynomial rings, there exists an eval homomorphism  that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$. 

Notice that the crucial insight is that $\;\rm  b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have *polynomial form* in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a *domain*, enjoys cancelation of elts $\ne 0$. 
Working generically allows us to cancel $\rm d$ and deduce the identity *before* any evaluation where $\rm d\mapsto 0.$ 

Such proofs by way of **universal polynomial identities** emphasize the power of the abstraction of a *formal* polynomial (vs. polynomial *function*). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.

Analogously, the same *generic* method of proof works for many other polynomial identities, e.g.

$\rm\quad\;  det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$

$\rm\quad\quad  det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$

Now, for our pièce de *résistance* (of limits, density...), we derive the polynomial derivative purely formally.

For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows 
from the  Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm  R[x,y],\;$ 
and, from the cancelation law  $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition 
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
*Resisting* limits again, we get the product rule rule for  derivatives from the trivial difference product rule 

$$ \rm f(x)g(x) - f(y)g(y)\;  =  \;(f(x)-f(y)) g(x)  +  f(y) (g(x)-g(y))$$

$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
 D(fg)\quad = \quad (Df) \; g \; +  \;  f \; (Dg) $

by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the *formal* cancelation of the factor $\;\rm x-y\;$  *before* evaluation at $\;\rm y = x\;$ is precisely analogous to the *formal* cancelation of $\rm det \:A$ in the example sparking this thread.