$-3x(8-x)^2 + (8-x)^3 = 0$ I've been asked to find the critical point of a function, where the first derivative is equal to $-3x(8-x)^2 + (8-x)^3$. In other words, I need to find all $x$ where $-3x(8-x)^2 + (8-x)^3 = 0$. 
However, I'm finding this algebra problem unexpectedly difficult. Online calculators have all solved for different values of x, which makes it difficult to confirm my solution. I would appreciate it if someone could give me a solution and/or review my solution below. Please outline your reasoning. If my reasoning is incorrect, please outline why. Thank you.
My solution is as follows.
$-3x(8 - x)^2 + (8 - x)^3 = 0$
$\Rightarrow$ $-3x(8 - x)^2 + (8 - x)^3 = (8 - x)^2\left[(8 - x) -3x\right]$
$\Rightarrow$ $-3x(8 - x)^2 + (8 - x)^3 = (8 - x)^2 (8 - 4x)$
$\therefore$ $x = 8, 2$ Are critical points

Yes, your solution is correct: 2 and 8 are the critical points, and your reasoning is clear.
