Are there any nicely factoring cubic equations with the same first three coefficients? For example is there a factorable equation of the form $(x+1)^3+c$ where $c\neq 0$, more generally is there a factorable equation of the form $(x+a)^{n}+c$ again with the condition that $c\neq0$, but also $n&gt;2$?
For $c \ne 0$ the equation $(x+a)^n + c = 0$ has exactly one real root if $n$ is odd, and two or no roots if $n$ is even and $c \lt 0\,$, $c \gt 0$ respectively. Therefore, there is no factorization possible in linear (real) factors for $n \ge 3$.
Here's a quick answer &mdash; just showing some possibilities, but certainly far from complete. Since $a^3+b^3=(a+b)(a^2-ab+b^2)$, pick $c$ to be a perfect cube, and you'll have a nicely factorable expression. The same trick works for all odd powers $n$.