<s>[INST] Is it possible to evaluate $232^2-62^2\times14$ by factoring or using identities or...? The expression $232^2-62^2\times14$ can be calculated directly ($53824-3844
\times14=8$). But is it possible to evaluate it for example by factoring or using identities?
Here is what I have tried,
$$(58\times4)^2-62^2\times14=58^2\times16-62^2\times14=(60-2)^2(15+1)-(60+2)^2(15-1)$$
Or
$$58^2\times16-62^2\times14=29^2\times64-31^2\times56=(30-1)^2\times8^2-(30+1)^2(8\times7)$$
But I can't see an elegant way to get $8$ from either of the calculatins.
 [/INST] The numbers involved are easy to factor, so take out the common factors to get
$$232^2-62^2\times14=(2^3\times29)^2-(2\times31)^2\times(2\times7)
=2^3\times(2^3\times29^2-7\times31^2).$$
The last factor looks similar to a difference of two squares; note that
\begin{eqnarray*}
2^3\times29^2-2^3\times31^2&=&2^3\times(29^2-31^2)\\
&=&2^3\times(29+31)\times(29-31)\\
&=&-2^4\times60\\
&=&-960,
\end{eqnarray*}
and so
$$2^3\times29^2-7\times31^2=-960+31^2=1.$$
It follows that
$$232^2-62^2\times14=2^3\times1=8.$$
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