**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 $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have *polynomial form* in $\;\rm a_{i,j}$, i.e. they are elt's 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 form many other identities, e.g.


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

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

Finally. for our piece de resistance (of limits!), we derive polynomial derivatives 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 $\rm x^n$.
Finally, resisting limits once again, we formally prove Leibniz's product rule rule for derivatives: 

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

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

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