given the function, f(x)=(x^3)-(3x^2)+(64x)-192 find the zeros

There are a few ways to find the zeros. I am going to use the grouping method.

NOTE: GCF = Greatest Common Factor

The function: [tex]f(x) = x^3 - 3x^2 + 64x - 192[/tex] 

Lets group [tex]x^3 - 3x^2[/tex] and [tex]64x - 192[/tex]

Factor [tex]x^3 - 3x^2[/tex]
Find GCF
GCF = [tex]x^2[/tex]

Divide out the GCF [tex]x^2[/tex]
[tex]\frac{x^3 - 3x^2}{x^2} = (x - 3)[/tex]

Factor From:
[tex]x^2(x - 3)[/tex]

Now factor 64x -192
GCF of 64 and 192 = 64
Factor out 64 from 64x - 192
[tex]\frac{64x - 192}{64} = (x - 3)[/tex]

Factor Form:
[tex]64(x - 3)[/tex]

Now that we have factored the groups, bring them together.
[tex] x^2(x-3)[/tex]
[tex] 64(x-3)[/tex]

[tex]x^2(x-3) + 64(x-3)[/tex]

Since (x-3) is the GCF of [tex]x^2(x-3) + 64(x-3)[/tex], factor (x-3) out.
[tex]\frac{x^2(x-3) + 64(x-3)}{(x-3)} = x^2 + 64[/tex]

Factor Form:
[tex](x-3)(x^2 + 64)[/tex]

Now set [tex](x^2 + 64)(x-3)[/tex] to zero and solve for x.
[tex](x^2 + 64) = 0[/tex]
[tex]x^2 = - 64[/tex]

Take the square root of both sides of the equal sign:
[tex]\sqrt{x^2} = \sqrt{-64}[/tex]
[tex]x = \sqrt{-64}[/tex]
[tex]x = \pm 8i[/tex]

Now for (x-3)
x - 3 = 0
x - 3 + 3 = 3
x = 3

The zeros and answers:
[tex]x = -8i[/tex]
[tex]x = 8i[/tex]
x = 3