Find all relative extrema and points of infelction

Answers are: Relative minimum: (1, -27), Points of infelction: (2, -16), (4, 0)

f(x)=x(x-4)^{3}

It could also be simplified into x^{4}-64x

f'(x)=3x(x-4) OR 4x^{3}-64

0=3x(x-4)^{2 }+ 3x OR 0=x^{4}-64x

Thus, x = 0, and x = 1

f''(x)=6x(x-4) + 3x + 6x OR 12x^{2}

f''(0)=6(0)(0-4) + 3(0) + 6(0)=0 so here is not a relative extrema

f''(3)=6(3)(3-4) + 3(3) + 6(3)=9(I think I think the product rule wrong for the original equation, and second derivative, if someone could help me fix that, so I'll use the other function instead to get it)f''(3)=12(1)

^{2 }= 1Here is the relative minima

So I plugged it back into the original equation and got the relative extrema to be (1, -27)

Now finding the Points of Inflection has been my problem

I went around and saw that people say to have the second derivative set equal to 0

0=12x^{2}

0=x

Then they say to plug that point into the original equation.

f(x) = 0(0-4)^{3}= 0

And that was not one of the points of inflection in the answer.

Can anyone explain? Thanks