Shortcut for finding cube of the Numbers Is there a shortcut for finding cube of a particular number like $68^3$ ? 
If anyone knows how to solve for two- and three digit numbers, can you please share the answer?
 A: You can apply the binomial theorem for a more comfortable calculation:
$$(a+b)^n=\sum_{k=0}^n~{n\choose k} \cdot a^{n-k} \cdot b^k$$
First note that $68=70-2$
Therefore in your case it is 
$$(70-2)^3=\sum_{k=0}^3~{3\choose k} \cdot 70^{3-k} \cdot (-2)^k$$
$=1\cdot 70^3\cdot 1+3\cdot 70^2\cdot (-2)+3\cdot 70\cdot (-2)^2+1\cdot  1 \cdot (-2)^3$
$=343,000-6\cdot 4900+12\cdot 70-8$
$=343,000-29.400+840-2=313,600+832=314,432$
In this case it can be calculated without using a calculator.
A: Trying to see how I think the following: if the number has small prime factors, you can first factoring them and then apply the binomial theorem  appropriately with respect to a multiple of 10, or calculate first the square of large prime factors. For example for your $68$ one can do
$$68=2^2\cdot 17\rightarrow68^3=2^6\cdot17^3=2^6\cdot17(20-3)^2$$
Perhaps for three digit numbers may be convenient to write instead of $abc$ the expression $$abc=10ab+c$$ after factorization (when it is easy of course, such as the $2^2$ for $68$) wich besides could lead to a two digit number such as 237 with its factor $3$.
