You may also use the difference of two cubes $\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)$ but Plato's method is better because of the factors involved when doing the difference of two cubes
A slight variation- if we let a= x- 1 then this is $\displaystyle (a+c)^2- a^3= a^3+ 3a^2c+ 3ac^2+ c^3- a^2= 3a^2c+ 3ac^2+ c^3$. Now replace a by x- 1 again to get
$\displaystyle 3c(a-1)^2+ 3c^2(a- 1)+ c^3$