Question 1094198
<br>I will guess that the "shortcut pattern" you refer to is from the 3rd row of Pascal's Triangle, which contains the numbers 1 3 3 1.  Those four numbers are the coefficients in the expansion of (a+b)^3.  More specifically, the expansion of (a+b)^3, using the 3rd row of Pascal's Triangle, is
{{{(a+b)^3 = (1)a^3 + (3)a^2b + (3)ab^2 + (1)b^3}}}<br>
To use the shortcut for cubing any binomial, just replace the a and b with the two terms in your binomial.  For example, if the binomial is (2x-5), then
{{{(2x-5)^3 = (2x+(-5))^3}}}
 = {{{(1)(2x)^3 + (3)(2x)^2(-5) + (3)(2x)(-5)^2 + (1)(-5)^3}}}
 = {{{8x^3 - 60x^2 + 150x - 125}}}<br>
Now try the same process with your binomial, (x+5).